Answer:
The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].
The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].
Explanation:
The oxidation states of atoms in a compound should add up to zero.
Ag₂OThere are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:
[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].
SO₂Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.
Therefore:
[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the oxidation state of [tex]\rm S[/tex] here:
[tex]\text{Oxidation state of $\rm S$} = 4[/tex].
An individual was injected with 80 mg of inulin and 960,000 counts per min (cpm) of tritium-labeled water (3H20) to determine the volume of various body fluid compartments. After equilibration a blood sample was obtained and the plasma inulin concentration was 0.5 mg% and the plasma activity (concentration) of tritium was 20 cpm/ml. The volumes of which body compartments can be determined?
The measurement of body fluid compartments can be achieved by the dilution of chemical compounds that only circulate and disperse in the region of selected areas in the body. The dilution process is dependent on how the concentration is defined.
Given that:
the concentration of plasma insulin after equilibrium = 0.5 mg %∴
Concentration C = 0.5 mg/100
Concentration C = 0.005 mg/ml
The mass of insulin = 80 mgSince the mass amount of the chemical compound(i.e. insulin) and the concentration is known.
The volume of the body fluid compartment can be calculated as:
[tex]\mathbf{volume = \dfrac{\text{mass of the marker }}{concentration }}[/tex]
[tex]Volume = \dfrac{80 \ mg}{0.005 \ mg/ml}[/tex]
Volume = 16000 ml
Thus, it is known that insulin is generally utilized for the measurement of the extracellular fluid volume and serves as a cell impermeant marker.
As a result;
The volume of the extracellular fluid compartment is 16000 ml.
However, the tritium-labeled water is a good marker for the entire body fluid compartment due to the fact that:
its diffusion occurs throughout the entire body,it is identical to water and;the equilibrium concentration is typically easy to measure due to the radioactive characteristics of tritium.Given that:
plasma activity of tritium = 20 cpm/ml
i.e.
In 1 ml of plasma, 20 cpm of tritium is present.
As such, in 960,000 counts per min (cpm) of tritium-labeled water, the volume of the whole body compartment is:
[tex]\mathbf{= \dfrac{960000}{20} ml \plasma}[/tex]
= 48000 ml of plasma
Therefore, we can conclude that the volumes of the body compartment that can be determined are:
The volume of the extracellular fluid compartment, which is 16000 ml.The volume of the whole body compartment, which is 48000 mlLearn more about body fluid compartments here:
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Consider the following reaction: 2Fe2+(aq) + Cu2+ --> 2Fe3+(aq) + Cu. When the ion concentrations change to the point where the reaction comes to equilibrium, what would be the cell voltage?
Answer:
At equilibrium, the cell voltage is zero volts.
Explanation:
During an electrochemical reaction, electrical energy is produced. The reaction continues to produce electrical energy until a point is reached in which the reaction attains equilibrium.
Before the reaction attains equilibrium, the cell voltage continues to decrease progressively as the reaction progresses. At equilibrium, the cell voltage becomes zero and the read out voltmeter records 0 V.
Hence, at equilibrium, the cell voltage is zero volts.
HELP ASAS 15 POINTS
When using the process of evaporation to separate a mixture, what is left behind in the evaporating dish?
A. None of these.
B. The liquid evaporates and the solid is left in the dish.
C. The mixture does not separate, and the entire mixture evaporates.
D. The mixture does not separate, and the entire mixture remains in the dish.
Answer:
liquid will be evaporated while solid remains
Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)
Explanation:
C5H10O2 + NaOH = C2H5COONa + C2H5OH
your result are : sodium propanoate and ethanol
A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.
What is hydrolysis?Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.
The base hydrolysis of methyl butanoate is shown as,
C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH
Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).
Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.
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what mass of aluminum nitrate do you need to prepare 3.58L of a 1.77M Solution?
Oxygen condenses into a liquid at approximately 90 K. What temperature, in degrees Fahrenheit, does this correspond to?
Answer:
-297.67 °F
Explanation:
Oxygen condenses into a liquid at approximately 90 K. We can convert any temperature in the Kelvin scale (absolute scale) to the Fahrenheit scale using the following expression.
°F = (K − 273.15) × 9/5 + 32
°F = (90 − 273.15) × 9/5 + 32
°F = (-183.15) × 9/5 + 32
°F = -329.67 + 32
°F = -297.67 °F
One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
g Use the References to access important values if needed for this question. A researcher took 2.592 g of a certain compound containing only carbon and hydrogen and burned it completely in pure oxygen. All the carbon was changed to 7.851 g of CO2, and all the hydrogen was changed to 4.018 g of H2O . What is the empirical formula of the original compound
Answer:
Empirical formula is: C₂H₅
Explanation:
The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:
CₓHₙ + O₂ → XCO₂ + n/2H₂O
That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:
Moles C and H:
Moles C = Moles CO₂:
7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C
Moles H = 2 Moles H₂O
4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H
Ratio C:H
The ratio between moles of hydrogen and moles of Carbon are:
0.4417 moles H / 0.1784 moles C = 2.5
That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,
Empirical formula is: C₂H₅Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2NO(g)+O2(g)⇌2NO2(g)
Part A If Kc=6.9×105 at 227 ∘C,
what is the value of Kp at this temperature? Express your answer using two significant figures. Kp =
Part BIf Kp=1.3×10−2 at 1000 K, what is the value of Kc at 1000 K? Express your answer using two significant figures. Kc =
Answer:
Kp=1.68×10⁴∆1.7×10⁴
Kc=1.06∆1.1
Explanation:
Value of Kp at 227°C is 2.86×10² and value of Kc at 1000 K is 1.56.
How are Kp and Kc related?Kp and Kc are related by the formula Kp=Kc(RT).For part A , Kp is calculated as,
Kp=6.9×10⁵×8.314×500=28.683×10² and for part B Kc is calculated as,
Kc=1.3×10[tex]^-2[/tex]/(8.314×1000)=1.56
Kc and Kp are equilibrium constants of a mixture of ideal gases. Kp is equilibrium constant when concentrations at equilibrium are in atmospheric pressure and Kc is equilibrium constant when concentrations are in molarity. The relation is only valid for gaseous mixtures. The relation between these two parameters is obtained through ideal gas equation.
Kc and Kp of reaction change with temperature of reaction but remain unaffected by change in concentration , pressure and presence of catalyst.
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A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
An HCl solution has a concentration of 0.09714 M. Then 10.00 mL of this solution was then diluted to 250.00 mL in a volumetric flask. The diluted solution was then used to titrate 250.0 mL of a saturated AgOH solution using methyl orange indicator to reach the endpoint.
Required:
a. What is the concentration of the diluted HCI solution?
b. If 7.93 mL of the diluted HCI solution was required to reach the endpoint, what is the concentration of OH- in solution?
c. What is the concentration of Ag+ in solution?
d. What is the Ksp expression for the dissolution of AgOH?
Answer:
a. 3.8856x10⁻³M HCl
b. 1.23x10⁻⁴M OH⁻
c. 1.23x10⁻⁴M Ag⁺
d. Ksp = [Ag⁺] [OH⁻]
Explanation:
a. The reaction that you are studying is:
HCl(aq) + AgOH(aq) → H₂O(l) + AgCl(s)
The HCl solution is diluted from 10.00mL to 250.00mL, that is:
250.00mL / 10.00mL = 25 -The solution is diluted 25 times-
As original concentration of HCl is 0.09714M, the concentration of the diluted solution is:
0.09714M / 25 =
3.8856x10⁻³M HClb. 1 mole of HCl reacts per mole of AgOH, moles of HCl that reacts are:
7.93mL = 7.93x10⁻³L × (3.8856x10⁻³mol HCl / L) = 3.0813x10⁻⁵ moles of HCl.
Based on the reaction, you have in solution
3.0813x10⁻⁵ moles of AgOH = Ag⁺ = OH⁻
The AgOH solution was 250.0mL = 0.2500L, its concentration is:
3.0813x10⁻⁵ moles OH⁻ / 0.2500L =
1.23x10⁻⁴M OH⁻c. In solution, AgOH produce Ag⁺ and OH⁻ in equals proportions, that means:
1.23x10⁻⁴M OH⁻ =
1.23x10⁻⁴M Ag⁺d. The solubility product reaction of AgOH(s) is:
AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)
Where Ksp for this reaction is defined as:
Ksp = [Ag⁺] [OH⁻]PLEASE HELP!!
this is on USAtestprep
a)
b)
c)
d)
How do covalent bonds form? A Sharing valence electrons between atoms. B Donating and receiving valence electrons between atoms. C Opposite slight charges attract each other between compounds. D Scientists are still not sure how they form.
Answer:
A. Sharing valence electrons between atoms.
Explanation:
This is the definition of a covalent bond. Option B describes ionic bonds, Option C describes intermolecular forces, and Option D is wrong because then there wouldn't be any mention of them in our high school chemistry textbooks :).
If you have 2.4L of SO2 gas (at STP) how many moles of sulfur dioxide do you have?
Answer:
0.107 mole of SO2.
Explanation:
1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).
With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.
This can be obtained as follow:
22.4 L contains 1 mole of SO2.
Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.
Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.
Please help due tomorrow
Answer:
OPTION C
Explanation:
AS IN HUMANS THE DNA COMES FROM BOTH MALE AND FEMALE-SPERM AND OVA . SO 23 CHROMOSOMES FROM FATHER AND 23 CHROMOSOME FROM MOTHER
How to do q solution, qrxn, moles of Mg , and delta Hrxn?
Answer:
14, 508J/K
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
m = mass of substance (g) = 0.1184g
1 mole of Mg - 24g
n moles - 0.1184g
n = 0.0049 moles.
Also, q = m × c × ΔT
Heat Capacity, C of MgCl2 = 71.09 J/(mol K)
∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)
= 14, 508 J/K/kg
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × 14, 508 j/K/kg
= 1,7117.7472 J °C-1 g-1
∴ ΔHrxn = q/n
=1,7117.7472 ÷ 0.1184
= 14, 508J/K
What is the compound formed from the combination of the base and a hydrogen ion
Answer:
Water
Explanation:
When a base react to and hydrogen ion, we can produce water.
According to these equation
H⁺ + OH⁻ ⇄ H₂O Kw: 1×10¹⁴
Remember that OH⁻ is determined by a strong base.
This reaction is called neutralization. You can also produce water with a weak base, because OH⁻ are released. For example, let's mention ammonia which is a weak base, it takes protons from water (H⁺)
NH₃ + H₂O ⇄ OH⁻ + NH₄⁺ Kb
When the ammonium ion (acid), reacts to a base, you produce water.
NH₄⁺ + NaOH → NH₃ + H₂O + Na⁺
A 1 L container originally holds 0.4 mol of N2, 0.1 mol of O2, and 0.08 mole of NO. If the volume of the container holding the equilibrium mixture of N2, O2, and NO is decreased to 0.5 L without changing the quantities of the gases present, how will their concentrations change
Answer:
a) [tex]N_2=0.8mol/L[/tex]
b) [tex]O_2=0.2mol/L[/tex]
c) [tex]NO=0.16mol/L[/tex]
Explanation:
From the question we are told that:
Moles 0f Nitrogen [tex]N_2=0.4[/tex]
Moles 0f Oxygen [tex]O_2=0.1[/tex]
Volume Decrease [tex]V_2=0.5L[/tex]
Generally, the equation for Concentration is mathematically given by
[tex]C=\frac{moles}{V}[/tex]
For Nitrogen
[tex]N_2=\frac{0.4}{0.5}[/tex]
[tex]N_2=0.8mol/L[/tex]
For Oxygen
[tex]O_2=\frac{0.1}{0.5}[/tex]
[tex]O_2=0.2mol/L[/tex]
For Nitrogen
[tex]NO=\frac{0.08}{0.5}[/tex]
[tex]NO=0.16mol/L[/tex]
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many
The given question is incomplete.
The complete question is:
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?
Answer: 4 grams of methane were needed for the reaction
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]
Given: mass of oxygen = 16 g
Mass of carbon dioxide = 11 g
Mass of water = 9 g
Mass of products = Mass of carbon dioxide + mass of water = 11 g +9 g = 20 g
Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g
As mass of reactants = mass of products
mass of methane + 16 g= 20 g
mass of methane = 4 g
Thus 4 grams of methane were needed for the reaction
The new hybrid car can get 51.5 km/gal. It has a top speed of 40000.00 cm/min and is 4m long. How fast can the car go in m/hr?
Answer:
The anawer of this question is 0.024 m/h
Explanation:
Other explanations of the question are additional.
Question 6
Which compound is more soluble in water?
O 3-propyl-2-octanol
1-methyl-3-pentanol
O2-methyl-1-ethanol
02-ethyl-3-heptanol
mouthing
Answer:
ethyl - 3- heptanol mouthing
How many moles of NaF must be dissolved in 1.00 liter of a saturated solution of PbF 2 at 25°C to reduce the [Pb 2+] to 1.0 × 10 –6 M? The K sp for PbF 2 at 25 °C is 4.0 × 10 –8.
Answer:
0.1957 moles of NaF
Explanation:
The Pb²⁺ and F⁻ are in equilibrium with PbF₂ as follows:
PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)
Where Ksp expression is:
Ksp = 4.0x10⁻⁸ = [Pb²⁺] [F⁻]²
A saturated solution contains the maximum possible amount of Pb²⁺ and F⁻. That is:
PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)
PbF₂(s) ⇄ X + 2X
Where X is amount of ions presents in solution
4.0x10⁻⁸ = [Pb²⁺] [F⁻]²
4.0x10⁻⁸ = [X] [2X]²
4.0x10⁻⁸ = 4X³
4.0x10⁻⁸/4 = X³
1.0x10⁻⁸ = X³
2.15x10⁻³M = X
That means initial concentration of Pb²⁺ is = X = 2.15x10⁻³M and [F⁻] = 2X = 4.30x10⁻³M
Now, using again Ksp, if you want a [Pb²⁺] = 1.0x10⁻⁶M, the [F⁻] you need is:
4.0x10⁻⁸ = [Pb²⁺] [F⁻]²
4.0x10⁻⁸ = [1.0x10⁻⁶M] [F⁻]²
0.04M = [F⁻]²
0.2M = [F⁻]
You need a final concentration of 0.2M of F⁻. As initial concentration was 4.30x10⁻³M and volume of the buffer is 1.00L, the moles of F⁻ = moles of NaF you must add are:
0.2M - 4.30x10⁻³M =
0.1957 moles of NaFIf you have a polyatomic anion of Ammonium (NH41+), how many valence electrons must your Lewis Structure have?
Answer:
One can draw the 3-dimensional structure of an atom once they have the Lewis Structure of an atom. The 3-dimensional geometrical structure of ammonium, NH4+ is referred to as Tetrahedral. ... But the + sign decrees that NH4+ has 8 valence shell electrons, due to the positive ion.
Explanation:
The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
H: 1 g/moleO: 16 g/molethe molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?
[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]
moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?
[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]
heat= 7.026 kJ
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
What is the relationship between concentration and rate of reaction?
not related
directly proportional
Inversely proportional
partially related
Explanation:
The answer is directly proportional, because when there is more concentration their will more reactants to react fast diring the chemical reaction which increases the rate of chemical reaction.
So, we can state that the relationship between them are directly proportional.
Hope it helps...
A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.
Answer:
[tex]\Delta G=-97.14kJ[/tex]
Explanation:
Hello,
In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]\Delta G=-RTln(K)[/tex]
Hence, we compute it as required:
[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]
And for 2.37 moles of hydrogen bromide, we obtain:
[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]
Best regards.
As a reaction proceeds, the ratio between the rate of consumption of reactant and the rate of formation of product:
Answer:
Depends on the reaction.
Explanation:
Hello,
In this case, the answer is depends on the reaction since the ratios between the rates of both consumption and formation depend upon the stoichiometric coefficients in the chemical reaction. For instance, for the reaction:
A -> 2B
The relationship is:
[tex]\frac{1}{-1}r_A =\frac{1}{2} r_B[/tex]
Therefore, we can see that the rate of consumption of A half the rate of formation of B, but is we consider the following chemical reaction:
2A -> B
The relationship is:
[tex]\frac{1}{-2}r_A =\frac{1}{1} r_B[/tex]
Therefore we can see that the rate of consumption of A doubles the rate of consumption of B.
Best regards.
Which processes occur during the second stage of technological design? Check all that apply.
designing a solution
studying relevant information
rebuilding and retesting
reporting a solution
defining criteria of success
identifying a problem
building a prototype
Calculate the energy required to heat of 1.50 kg silver from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.
Answer:
804 J
Explanation:
Step 1: Given data
Mass of silver (m): 1.50 kgInitial temperature: -7.8 °CFinal temperature: 15.0 °CSpecific heat capacity of silver (c): 0.0235J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]
Q = 804 J