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2 years ago

Which of the following statements is true about the following reaction?

Chemistry

2 answers:

MAVERICK [17]2 years ago

8 0

The correct reaction equation is:

$3NaHCO_{3} (aq) + C_{6}H_{8}O_{7} (aq) \rightarrow 3CO_{2} (g) + 3H_{2}O (l) + Na_{3}C_{6}H_{5}O_{7} (aq)$

Answer:

b) 1 mole of water is produced for every mole of carbon dioxide produced.

Explanation: CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.

a) 22.4 L of $CO_{2}$ gas is produced only when $\frac{22.4}{3}$ L of $C_{6}H_{8}O_{7}$ is reacted with 22.4 L of $NaHCO_{3}$ . So it is wrong.

b) Since in the chemical equation the stoichiometric coefficient of $CO_{2}$ and $H_{2}O$ are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. Therefore it is correct option.

c) $6.02\times 10^{23}$ molecules is equal 1 mole of $Na_{3}C_{6}H_{5}O_{7}$ if produced then 3 moles of $NaHCO_{3}$ is required, which is not given in the option. So it is wrong.

d) 54 g of water or 3 moles of $H_{2}O$ (Molecular Weight of water is 18 g) is produced when 3 moles of $NaHCO_{3}$ is used but in this option only one mole of $NaHCO_{3}$ is given. So it is wrong.

PIT_PIT [208]2 years ago

7 0

Answer:

Connexus: A,A,A,B.

Explanation:

These are the correct answers to Chemical Calculations Lesson 2 unit 6 stoichiometry answers.

Hope this helps:)

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Which nuclear emission moving through an electric field would be deflected toward the positive electrode?

Tom [10]

As a rule, opposite charges attract. So,you would expect that the particle that would deflect towards the positive electrode has a negative charge. Now, an alpha particle is basically a neutral Helium atom. Beta particle is an electron. Gamma particle has no charge (neutral). Thus, the answer is B.

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3 years ago

A gas has a pressure of 3.16 atm at STP. I have decided to transfer it to a container that is 3 times larger than the original v

Maru [420]

Answer: The new pressure will be 1.1atm

Explanation:

V1 = V

P1 = 3.16atm

V2 = 3V

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P2 = (3.16 x V) /3V

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2 years ago

PLEASE HELP!!! The image represents the reaction between a certain number of molecules of N2 and H2.

77julia77 [94]

Answer:

The leftover reactant is the nitrogen gas, N₂.

Explanation:

As per your description:

1. Square on the left: N₂(g)

3 units of two joint circles: this represents 3 molecules of nitrogen gas, N₂(g).

2. Square on the right: H₂(g)

3 units of two joint circles: this represents 3 molecules of hydrogen gas, H₂(g).

3. Reaction

If the maximum possible amount of NH₃ is formed during the reaction, you assume that the reaction goes to completion.

The chemical equation that represents the reaction is:

N₂(g) + H₂(g) → NH₃(g)

Which must be balanced:

N₂(g) + 3H₂(g) → 2NH₃(g)

That means that 1 molecule (or 1 mol) of N₂(g) reacts with 3 molecules (or 3 moles ) of H₂(g) to produce 2 molecules (or 2 moles) of NH₃(g).

Since, the squares show that there are 3 molecules of each reactant, the 3 molecules of hydrogen gas will be able to react with 1 molecule of nitrogen gas. When that happens, all the hydrogen gas is consumend and yet two molecules of nitrogen gas will remain unreacted. Hence, the nitrogen gas is the leftover reactant.

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2 years ago

Read 2 more answers

A molecule of chlorine has atoms of the same element, and a molecule of carbon dioxide has different kinds of atoms.based on thi

babymother [125]

Chlorine is an element and carbon dioxide is a compound

7 0

2 years ago

Read 2 more answers

What is the change in enthalpy in kilojoules when 3.24 g of CH3OH is completely reacted according to the following reaction 2 CH

vodka [1.7K]

Answer:

12.78 kJ

Explanation:

The correct balanced reaction would be

$2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}$

Mass of methanol = $3.24\ \text{g}$

Moles of methanol can be obtained by dividing the mass of methanol with its molar mass $(32.04\ \text{g/mol})$

$\dfrac{3.24}{32.04}=0.10112\ \text{moles}$

Enthalpy change for the number of moles is given by

$\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}$

$\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}$

The change in enthalpy is 12.78 kJ.

5 0

2 years ago