Answer:
Changes in environment, food source changes and disease?
Explanation:
I dont know what the answer choices are
Answer:
changes in the environment
Explanation:
like digging big holes.
Hope this helps!
Measurements show that unknown compound X has the following composition: element mass % carbon 41.0% hydrogen 4.58% oxygen 54.6% Write the empirical chemical formula of X.
Answer:
CHO
Explanation:
Carbon = 41%, Hydrogen = 4.58%, oxygen = 54.6%
Step 1:
Divide through by their respective relative atomic masses
41/ 12, 4.58/1, 54.6/16
3.41 4.58 3.41
Step 2:
Divide by the lowest ratio:
3.41/3.41, 4.58/3.41, 3.41/3.41
1, 1, 1
Hence the empirical formula is CHO
Answer:
The empirical formula of X is C3H4O3.
Explanation:
Which is one way that minerals crystallize from materials dissolved in water?
from the air
from solutions that evaporate
from hot water solutions when water boils
from the soil
Answer:
the second answer its science behind it
Answer:
b
Explanation:
4. Horizontal rows of the Periodic Table are called:
a, Clusters
Groups
b. Families
d) Periods
Consider the diagram below.
What does C represent?
A) enthalpy of reaction
B) activation energy
C) activated complex
D) energy of the reactants
Answer:
A) enthalpy of reaction
Explanation:
The region C signifies the enthalpy of reaction.
This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.
Enthalpy changes are heat changes accompanying a physical and chemical change. An enthalpy is the difference between the sum of the heat contents of products and sum of the heat contents of reactants.it is indeed A) enthalpy of reaction
SOMEONE PLZ HELP!!!!
Answer:
4.22mL
Explanation:
V=m/d
v= 18.45g/4.37g/mL
What concentration of NO−3NO3− results when 897 mL897 mL of 0.497 M NaNO30.497 M NaNO3 is mixed with 813 mL813 mL of 0.341 M Ca(NO3)2?
Answer:
Explanation:
NaNO₃ = Na⁺ + NO₃⁻¹
.497 M .497 M
moles of NO₃⁻¹ = .897 x .497 = .4458 moles
Ca( NO₃)₂ = Ca + 2 NO₃⁻¹
.341 M 2 x .341 M = .682 M
moles of NO₃⁻¹ = .813 x .682 = .5544 moles
Total moles = .4458 moles + .5544 moles
= 1.0002 moles
volume of solution = 897 + 813 = 1710 mL
= 1.710 L
concentration of nitrate ion = 1.0002 / 1.710 M
= .585 M
Use the Rydberg Equation to calculate the energy in Joules of the transition between n = 7 and n = 3 for the hydrogen atom. Find the frequency in Hz of this transition if the wavelength is 1000nm.
Answer:
The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].
The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].
Explanation:
The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.
Let [tex]\lambda_\text{vac}[/tex] denote the wavelength of the photon released when measured in vacuum.Let [tex]R_\text{H}[/tex] denote the Rydberg constant for hydrogen. [tex]R_\text{H} \approx 1.09678 \times 10^{7}\; \rm m^{-1}[/tex].Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the principal quantum number of the initial and final main energy level of that electron. (Both [tex]n_1\![/tex] and [tex]n_2\![/tex] should be positive integers; [tex]n_1 > n_2[/tex].)The Rydberg Equation gives the following relation:
[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].
Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:
[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].
In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:
[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].
Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].
Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:
[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].
Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].
Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:
[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].
The energy of the transition between n = 7 and n = 3 is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.
Using the Rydberg Equation for energy;
ΔE = -RH(1/n^2final - 1/n^2initial)
Given that;
nfinal = 3
ninitial = 7
RH = 2.18 × 10^-18 J
ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)
ΔE = - 2.18 × 10^-18(0.11 - 0.02)
ΔE = - 1.96 × 10^-19 J
For the second part;
Since the wavelength is 1000nm, we have;
λ = 1000nm
c = 3 × 10^8 m/s
f = ?
c = λf
f = c/λ
f = 3 × 10^8 m/s/1000 × 10^-9 m
f = 3 × 10^8 m/s/ 1 × 10^-6 m
f = 3 × 10^14 Hz
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Which of the following best describes the structure of a nucleic acid?
a
Carbon ring(s)
b
Globular or fibrous
c
Single or double helix
d
Hydrocarbon(s)
Which 2 main body systems work alongside the digestive system?
Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with potassium hydroxide. It requires 22.9 mL of 1.430 M KOH solution to titrate both acidic protons in 54.2 mL of the carbonic acid solution.
Required:
a. Write a balanced net ionic equation for the neutralization reaction. Include physical states.
b. Calculate the molarity of the carbonic acid solution.
Answer:
a. H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)
b. 0.603 M
Explanation:
Step 1: Write the neutralization reaction
H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)
Step 2: Calculate the reacting moles of KOH
22.9 mL of 1.430 M KOH react.
0.0229 L × (1.430 mol/L) = 0.0327 mol
Step 3: Calculate the reacting moles of H₂CO₃
The molar ratio of H₂CO₃ to KOH is 1:1. The reacting moles of H₂CO₃ are 1/1 × 0.0327 mol = 0.0327 mol.
Step 4: Calculate the molarity of H₂CO₃
0.0327 moles of H₂CO₃ are in a volume of 54.2 mL. The molarity of H₂CO₃ is:
M = 0.0327 mol/0.0542 L = 0.603 M
Nicotinic acid, HC6H4NO2, is a B vitamin. It is also a weak acid with Ka = 1.4 × 10-5. Calculate [H+] and the pH of a 0.041 M solution of HC6H4NO2.
Answer:
[H+] = 7.576x10⁻⁴M
pH = 3.12
Explanation:
Based on the equilibrium of the nicotinic acid in water:
HC6H4NO2(aq) + H2O(l) ⇄ C6H4NO2-(aq) + H3O+(aq)
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
As both C6H4NO2-(aq) and H3O+(aq) comes from the same equilibrium, we can approximate their concentration as X and replace:
Ka = [C6H4NO2-] [H3O+] / [HC6H4NO2]
1.4x10⁻⁵ = [X] [X] / [0.041M]
5.47x10⁻⁷ = X²
7.576x10⁻⁴M = X = [H+]And as pH is defined as -log [H⁺]
pH = 3.12An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 5.50 g of this compound produced 8.07 g of carbon dioxide and 3.30 g of water.
Required:
a. How many moles of carbon, C, were in the original sample?
b. How many moles of hydrogen, H, were in the original sample?
Answer:
a. 0.183 mol C
b. 0.366 mol H
Explanation:
Assuming total combustion, all of the carbon in the unknown compound turned into carbon dioxide, CO₂.
So first we calculate the CO₂ moles produced, using its molecular weight:
8.07 g CO₂ ÷ 44 g/mol = 0.183 mol CO₂This means in the unknown compound there were 0.183 moles of carbon, C.
Conversely, all of the hydrogen in the unknown compound turned into water, H₂O.
Calculating the H₂O moles:
3.30 g ÷ 18 g/mol = 0.183 mol H₂OWe multiply the water moles by two, as there are 2 H moles per H₂O mol:
0.183 * 2 = 0.366 mol H.You have a carbonate buffer with pH 10.3 and a concentration of 2.0 M. What is the buffer capacity of 100 mL of the buffer against 3.0 M CsOH?
Answer:
Explanation:
pH = 10.3
[ H] = 10⁻¹⁰°³
= 5 x 10⁻¹¹ M
concentration of CsOH C = 3 M
pKa of carbonate = 6.35
Ka = 10⁻⁶°³⁵ = 4.46 x 10⁻⁷
Buffer capacity = 2.303 x C x Ka x [ H⁺] / ( Ka + [ H⁺]² )²
= 2.303 x 3 x 4.46 x 10⁻⁷ x 5 x 10⁻¹¹ / ( 4.46 x 10⁻⁷ + 25 x 10⁻²² )²
= 154 x 10⁻¹⁸ / 19.9 x 10⁻¹⁴
= 7.74 x 10⁻⁴ .
Josh heated a certain amount of blue copper sulfate crystals to get 2.1 g of white copper sulfate powder and 1.4 g of water. What is most likely the mass of the blue copper sulfate that he heated and why?
Answer: The mass of blue copper sulfate is 3.5 g
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the heating of copper sulfate crystals is:
Let the mass of blue copper sulfate be 'x' grams
We are given:
Mass of copper sulfate powder = 2.1 grams
Mass of water = 1.4 grams
Total mass on reactant side = x
Total mass on product side = (2.1 + 1.4) g
So, by applying law of conservation of mass, we get:
Hence, the mass of blue copper sulfate is 3.5 grams
Which profile best shows the topography alone line AD
20 POINTS PLEASE ANSWER ASAP!!!!
Why is calcium (Ca) in group 2, period 4 on the periodic table?
A Calcium, like all group 2 elements, is nonreactive and a gas at room temperature.
B Calcium, like all period 4 elements, is nonreactive and a gas at room temperature,
C Calcium, like all group 2 elements, is reactive and a solid at room temperature.
D. Calcium, like all period 4 elements, is reactive and a solid at room temperature.
Calcium (Ca) is in group 2 and period 4 on the periodic table be because Calcium has 2 valence electrons and 4 electron shell. Thus, calcium is a metal like all other group 2 element.
The correct answer to the question is Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.
Calcium is a group 2 element majorly because it has 2 valence electrons. It is also in period 4 because it has 4 electron shells.
Being a group 2 element, calcium is a solid at room temperature and also reactive. All elements in the group 2 are metals.
There are other elements in period 4 which are not solid. For example krypton is an element in period 4 and it is a gas and not reactive.
From the above information, we can conclude that the correct answer to the question is:
Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.
See attached image
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A certain chemical reaction releases of heat for each gram of reactant consumed. How can you calculate the heat produced by the consumption of of reactant? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Complete Question
The complete question is shown in the first uploaded image
Answer:
So the math expression is
[tex]heat = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
Explanation:
From the question we are told that
The heat released for 1 gram of reactant consumed is [tex]H = 37.5 \ KJ/g [/tex]
The mass of reactant considered is [tex]m = 1.9 \ kg = 1900 \ g[/tex]
So if
[tex]37.5 \ KJ [/tex] is produced for 1 gram
Then
x kJ is produced for 1900 g
=> [tex]x = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
So the heat released is
[tex]heat = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
Is nuclear fission exothermic or endothermic? Explain your answer.
Answer:
Exothermic
Explanation:
Nuclear fission means splitting, so there is a lot of energy being released.
Help, 8th grade Science
How many liters of H2(g) at STP are produced per gram of Al(s) consumed in the following reaction? 2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)
Answer:
1.24 L of H₂ at STP .
Explanation:
2Al(s) +6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
2 moles 3 x 22.4 L
2 x 27 g of Al reacts to give 3 x 22.4 L of H₂ at STP .
1 g of Al will react to give 3 x 22.4 / ( 2 x 27 ) L of H₂ at STP .
= 1.24 L of H₂ at STP .
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
The balanced chemical reaction has been given as:
[tex]\rm 2\;Al\;+\;6\;HCl\;\rightarrow\;2\;AlCl_3\;+\;3\;H_2[/tex]
From the equation, 2 moles of Aluminum gives 3 moles of Hydrogen
The mass of the compound from moles can be given as:
Mass = moles × molecular mass
Mass of 2 moles Al = 2 × 27 g
Mass of 2 moles Al = 54 g
Mass of 3 moles hydrogen = 3 × 2 g
Mass of 3 moles hydrogen = 6 g
From the equation,
54 g aluminum gives = 6 grams hydrogen
[tex]\rm 1\;gram\;aluminum\;=\;\dfrac{6}{54}\;\times\;1[/tex]
1 gram Aluminum = 0.11 grams hydrogen
The mass of hydrogen produced by 1 gram Al has been 0.11g. The moles equivalent to 0.11g hydrogen has been given as:
Mass = moles × molecular mass
0.11 g = moles × 2 g/mol
Mole of hydrogen = 0.055 mol
The moles of hydrogen produced by 1 gram of Al has been 0.055 mol.
According to the ideal gas equation, any gas at STP has 1 mole equivalent to 22.4 L. So,
1 mol = 22.4 L
0.055 mol = 0.055 × 22.4 L
0.055 mol = 1.244 L.
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
For more information about volume at STP, refer to the link:
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What is the momentum of a 1kg ball moving at 5m/s?
Answer:
5Ns
momentum= mass *velocity
=1*5
=5Ns
The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga
Answer:
2.35 x 10²⁰ atoms Ga
Explanation:
After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.
[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]
Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.
The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
StoichiometryFrom the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.
First, we will determine the number of moles of Ga present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]
Mass = 27.2 mg = 0.0272 g
Molar mass = 69.72 g/mol
Then,
[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]
[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles
Now, for the number of atoms present
From the formula
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of Ga atoms = 0.000390132 × 6.022×10²³
Number of Ga atoms = 2.35 × 10²⁰ atoms
Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
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Problem:
[Ar]4s2
Identify the period (p) , group (g) and valence electrons block of the element
Answer:
it is Calcium (Ca)
4th period, 2nd group, 2 valence electrons
How many moles of H2 are needed to produce 24 moles of NH3?
Answer:
36 mol of H2
Explanation:
The balanced equation of the reaction is given as;
3H2 + N2 --> 2NH3
From the reaction;
It takes 3 mol of H2 reacting with 1 mol of N2 to form 2 mol of NH3
3 mol of H2 = 2 mol of NH3
x mol of H2 = 24 mol of NH3
x = (24 * 3) / 2 = 36 mol of H2
plz help answer both will mark brainest
3. A certain Chemical Industry company has a quality control job opening. The job is open for any major with basic knowledge of chemistry. You decided to apply. In the interview the HR personnel gives you a sealed folder from a certain lot to test your laboratory experience, as well as your quantitative and volumetric analysis skills. The chemical contained in the sample is benzoic acid (C-H602) and it is known to be a monoprotic acid. In order to get the job, you need to determine if the sample's purity is acceptable based on their standards. Inside the folder you found a vial with a solid sample labeled BA-I, a periodic table, and the following data: 1.250 g of the sample required 20.15 mL of 0.500 M concentration of NaOH to reach the end point. The lot can be denied if the purity is below 99.5 % purity.
1) What is the purity in the sample?
2) Is it the purity acceptable?
3) Would you repeat the titration experiment?
Answer:
1) 97.6%
2) No the purity is not acceptable because the standard is 99.5% purity.
3) Yes I will repeat the titration experiment to confirm my result.
Explanation:
Equation of the reaction;
C7H6O2(aq) + NaOH(aq) ---------> C7H5ONa(aq) + H2O(aq)
From the information provided;
Number of moles of NaOH reacted = concentration × volume = 20.15/1000 × 0.500 = 0.01 moles
From the reaction equation;
1 mole of C7H6O2 reacts with 1 mole of NaOH
Hence 0.01 moles of C7H6O2 will react with 0.01 moles of NaOH
Mass of C7H6O2 reacted = number of moles of C7H6O2 × molar mass of C7H6O2
Molar mass of C7H6O2 = 122.12 g/mol
Mass of C7H6O2 reacted = 0.01 moles × 122.12 g/mol = 1.22 g
Percentage by mass of pure C7H6O2 in the impure sample = 1.22/1.250 × 100 = 97.6 %
PLS PLS PLS PLS PLS PLS PLS help ASAP!!!!! Scientists call all of the compounds that contain carbon and are found in living things Organic because ________.
WILL DO BRAINLIEST!
Determine each type of reaction. 2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l ) 2CX2HX2(g)+5OX2(g)⟶4COX2(g)+2HX2O(l) Choose... N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l ) NHX4NOX3(s)⟶NX2O(g)+2HX2O(l) Choose... C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) CO(g)+2HX2(g)⟶CHX3OH(l) Choose... 2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g ) 2Fe(s)+6HCl(aq)⟶2FeClX3(aq)+3HX2(g) Choose... C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s ) CaClX2(aq)+NaX2COX3(aq)⟶2NaCl(aq)+CaCOX3(s) Choose...
Answer:
2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction
N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction
C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction
2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction
C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction
Explanation:
We can determine the type of reaction by considering the reactants and products.
Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.
A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.
A combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form methanol.
An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).
A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.
Reactions are the formation of the products from the reactant. The types of reactions are combustion, decomposition, combination, Redox and double displacement.
What are the types of reactions?The reaction is a chemical change in the properties of the reactant that forms the products. It can be of various types based on the formation of the product.
The first reaction is combustion as the reactants react and use oxygen to form heat, carbon dioxide and water. The combustion reaction of ethyne can be shown as,
[tex]\rm 2 C _{2} H _{2} ( g ) + 5 O _{2} ( g ) \rightarrow 4 C O _{2} ( g ) + 2 H _{2} O ( l )[/tex]
The second reaction is decomposition in which a single reactant decomposes to form two or more products. The decomposition of ammonium nitrate can be shown as,
[tex]\rm N H _{4} N O _{3} ( s ) \rightarrow N _{2} O ( g ) + 2 H _{2} O ( l )[/tex]
The third reaction is a combination reaction in which two compound or elements combines to form one product. The combination reaction between carbon monoxide and hydrogen to form methanol can be shown as,
[tex]\rm C O ( g ) + 2 H _{2} ( g ) \rightarrow C H _{3} O H ( l )[/tex]
The fourth reaction is redox and includes the oxidation and the reduction of the species of the reaction. In the reaction, iron undergoes oxidation and hydrogen reduction. The redox reaction can be shown as,
[tex]\rm 2 F e ( s ) + 6 H C l ( a q ) \rightarrow 2 F e C l _{3} ( a q ) + 3 H _{2} ( g )[/tex]
The fifth reaction is a double displacement reaction in which the calcium and sodium interchange their position in the product formation. The reaction can be shown as,
[tex]\rm C a C l _{2} ( a q ) + N a _{2} C O _{3} ( a q ) \rightarrow 2 N a C l ( a q ) + C a C O _{3} ( s )[/tex]
Therefore, the type of reactions is 1. combustion, 2. decomposition, 3. combination, 4. redox and 5. double displacement.
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Question 5 of 5
Which two phrases describe the nature of an electromagnetic force?
O A. Acts only when objects touch each other
B. Produced by interactions between magnetic objects
O c. Not a fundamental force of nature
O D. Produced by interactions between electrically charged objects
Answer:
I think it's A and D
Explanation:
I'm not sure if it's right
Answer:
The answer is B and D
Explanation:
trust fr
A major component of gasoline is octane (C8H18). When octane is burned in air, it chemically reacts with oxygen gas (O2) to produce carbon dioxide CO2 and water H2O. What mass of carbon dioxide is produced by the reaction of 7.58 g of octane? Please explain the answer to me like I'm five, I want to understand but the content makes no sense.
Answer:
Mass = 23.232 g
Explanation:
Given data:
Mass of C₈H₁₈ = 7.58 g
Mass of CO₂ produced = ?
Solution:
Chemical equation:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
Number of moles of octane:
Number of moles = mass/molar mass
Number of moles = 7.58 g/ 114.23 g/mol
Number of moles = 0.066 mol
Now we will compare the moles of CO₂ with octane from balance chemical equation.
C₈H₁₈ : CO₂
2 : 16
0.066 : 16/2×0.066 = 0.528
Mass of CO₂ produced:
Mass = number of moles × molar mass
Mass = 0.528 mol × 44 g/mol
Mass = 23.232 g