If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?

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

Answer:

Chromosome condensation, the landmark event at the onset of prophase, often begins in isolated patches of chromatin at the nuclear periphery. Later, chromosome condense into two threads termed sister chromatids that are closely paired along their entire lengths.

Explanation:

hope dis help & may I have brainly plz

Answer:

C. water acts as a radiation shield to reduce the radiation level

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:

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

Answer:

option c see image... and yw

The molecule shown in option C is butane.

To learn more about butane,

https://brainly.com/question/14818671

#SPJ2

Answer:

For the mentioned reaction, the balanced chemical equation is:  

KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)

The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.  

From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.  

Answer:

B. 4.52 X10-9 M

Explanation:

Our goal for this question is to calculate the concentration of hydronium ions [tex]H^+[/tex] produced by water in a vessel with a concentration of hydroxide ions of [tex]2.21X10^-^6~M[/tex]. So, our first approach can be the ionization reaction of water:

[tex]H_2O_(_l_)~->~H^+~_(_a_q_)~+~OH^-~_(_a_q_)[/tex]

If we write the Keq expression for this reaction we will have:

[tex]Keq=[H^+][OH^-][/tex]

Now, water is the universal solvent, so, Keq has a special name. In the equilibrium problems for water we have to use "Kw" instead of "Keq":

[tex]Kw=[H^+][OH^-][/tex]

From this equation, we know the Kw value () and the concentration of the hydroxide ions ([2.21X10^-^6~M]). If we replace these values into the equation we can solve for [tex][H^+][/tex]:

[tex]1.0X10^-^1^4=[H^+][2.21X10^-^6~M][/tex]

[tex][H^+]=\frac{1.0X10^-^1^4}{2.21X10^-^6}=4.52X^-^9[/tex]

I hope it helps!

Answer:

B. 4.52 * 10^-9M

Explanation:

did the test

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

Answer:

6.93

Explanation:

Step 1: Given data

Step 2: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 50°C + 273.15

K = 323 K

Step 3: Calculate K

We will use the following expression.

∆G° = -R × T × ln K

-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K

K = 6.93

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:

the 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.

Answer:

Anaphase

Explanation:

The centromere splits during the anaphase of the cell division. Thus, allowing the two linked chromatids to separate.

A typical chromosome is made up of two sister chromatids joined together by a structure known as the centromere. During cell division - at the metaphase stage - the chromosomes align at the equator of the cell, forming the metaphase plate. The spindle from the opposing ends of the cell engages each chromosome at the kinetochore of the centromere.

At the anaphase stage, the centromere splits, leading to the separation of the sister chromatids of each chromosome. The sister chromatids of the same then start migrating in the opposite direction as a result of the shortening of the spindle fiber.

Answer:

4.22

Explanation:

We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].

If the pOH of a solution is given, one may obtain the pH of such solution from the formula;

pH + pOH =14

Hence we can write;

pH = 14-pOH

pH = 14 - 9.78 = 4.22

Hence the pH of the solution is 4.22.

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.

Answer:

AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

Explanation:

If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq)  with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.

The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of  silver trioxonitrate(V) is added afterwards. A  white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.

The reaction proceeds as follows:

[tex]\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}[/tex]

From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

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.

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.

Learn more about equilibrium constants,here:

https://brainly.com/question/15118952

#SPJ2

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,

Answer: A. [tex]H_{2}_{(g)}+I_{2}_{(g)}=>2HI_{(g)}[/tex] and D.[tex]C_{(s)}+O_{2}_{(g)}=>CO_{2}_{(g)}[/tex]

Explanation: The relationship between internal energy change and enthalpy change during a chemical reaction occurs according to the following formula:

[tex]\Delta H=\Delta E+\Delta(PV)[/tex]

So, for changes in enthalpy and internal energy to be equal volume or pressure has to be constant, i.e., zero.

Change in the number of moles of gas during the reaction can make the difference between [tex]\Delta H[/tex] and [tex]\Delta E[/tex] be larger, so for them to be equal and pressure constant, number of moles must be the same in reagents and products.

Analysing each reaction above:

Reaction A has the same number of moles in reagents and products, so  enthalpy change and internal energy change will be equal;

Reactions B and C don't have the same number of moles at both sides, so enthalpy and energy will be different.

Reaction D, although reagent side have 2 compounds, carbon is solid, so reaction have the same number of moles in both sides. Enthalpy and Energy will be equal.

Answer:

The correct answer is 206 ml.

Explanation:

Based on the given information, the molarity or M₁ of MgF₂ solution is 0.0887 M, the molarity or M₂ of the final solution given is 0.0224 M. The initial volume of V₁ of the solution is 69.6 ml, for finding the final volume of V₂ of the solution, the formula to be used is,  

M₁V₁ = M₂V₂

Now putting the values in the formula we get,  

0.0887 × 69.6 = 0.0224 M × V₂

V₂ = 0.0887 × 69.6 / 0.0224

V₂ = 275.6 ml

Therefore, the volume in ml added to the initial volume of 69.6 ml to make the molarity of the solution 0.0224 will be,  

= 275.6 ml - 69.6 ml = 206 ml

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

Answer:

ethyl - 3- heptanol mouthing

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 =

b. 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 =

c. In solution, AgOH produce Ag⁺ and OH⁻ in equals proportions, that means:

1.23x10⁻⁴M OH⁻ =

d. The solubility product reaction of AgOH(s) is:

AgOH(s) ⇄ Ag⁺(aq) + OH⁻(aq)

Where Ksp for this reaction is defined as:

Answer:

liquid will be evaporated while solid remains

Answer:

The anawer of this question is 0.024 m/h

Explanation:

Other explanations of the question are additional.

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.

Answer:

TATGGCGTT

Explanation:

Complimentary base pairs:

A-T

C-G

Use the other letter for complimentary strands

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...

Answer:

The answer is

Explanation:

The Ka of an acid when given the pH and concentration can be found by

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

Multiply through by - 2

Find antilog of both sides

We have the final answer as

Hope this helps you