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2 years ago

What are the coefficients that would balance this equation _NaOH+H2CO3-Na2CO3+_H2O

Chemistry

1 answer:

FromTheMoon [43]2 years ago

4 0

I don’t see any equal signs to make it an equation. Am I missing something?

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2NO2 --> N2 + 2O2

kow [346]

C. Decomposition

Since one reactant turns (decomposes) into two products, it is a decomposition reaction

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3 years ago

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If the azeotropic mixture is 72% perchloric acid by mass, what is the mole percent of water in the azeotrope?

WARRIOR [948]

Answer:

Mole percent of water in azeotrope is 69%

Explanation:

Clearly, the azeotrope consists of water and perchloric acid.

So, 72% perchloric acid by mass means 100 g of azeotrope contains 72 g of perchloric acid and 28 g of water.

Molar mass of water = 18.02 g/mol and molar mass of perchloric acid = 100.46 g/mol

So, 72 g of perchloric acid = $\frac{72}{100.46}mol$ of perchloric acid = 0.72 mol of perchloric acid

Also, 28 g of water = $\frac{28}{18.02}mol$ of water = 1.6 mol of water

Hence total number of mol in azeotrope = (1.6+0.72) mol = 2.32 mol of azeotrope

So, mole percent of water in azeotrope = [(moles of water)/(total no of moles)] $\times 100$ %

Mole percent of water = $\frac{1.6}{2.32}\times 100$ % = 69%

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4 years ago

How you think and feel affects the way you behave.<br> True<br> False

Vika [28.1K]

Answer:

True

Explanation:

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3 years ago

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Using the Fick equation, calculate cardiac output if a person is consuming 1.41 L O2/min, has an arterial O2 content of 190 ml O

Elden [556K]

Answer:

Cardiac output (CO) = 8.54 l / min

Explanation:

Given:

Oxygen consumption (V) = 1.41 l O2/min

Oxygen in arterial (Ca) = 190 ml O2/L = 0.19 l O2/L

Oxygen in venous (Cv) = 25 ml O2/L = 0.025 l O2/L

Find:

Cardiac output (CO)

Computation:

Cardiac output (CO) = V / (Ca - Cv)

Cardiac output (CO) = 1.41 / (0.19 - 0.025)

Cardiac output (CO) = 8.54 l / min

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3 years ago

The heat capacity of benzene (c6h6) is 81.7 jmol–1°c–1. how much heat (in joules) is required to raise the temperature of 39.0 g

Viefleur [7K]

Answer is: .408,5 J.m(C₆H₆-benzene) = 39 g.
n(C₆H₆) = m(C₆H₆) / M(C₆H₆) = 39g ÷ 78g/mol = 0,5 mol.ΔT = 10,0°C, difference at temperature.c(benzene) = 81,7 J/mol·°C, specific heat of benzene.Q = n(benzene) · ΔT · c(benzene), heat of reaction.Q = 0,5 mol · 10,0°C · 81,7 J/mol·°C.Q = 408,5 J.

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3 years ago

What are the coefficients that would balance this equation _____NaOH+____H2CO3-Na2CO3+_____H2O

What are the coefficients that would balance this equation _NaOH+H2CO3-Na2CO3+_H2O