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3 years ago

Which option is a balanced equation for cellular respiration?

Chemistry

2 answers:

qwelly [4]3 years ago

8 0

The first one, but should be:
C6H12O6 + 6O2 -> 6CO2 + 6 H2O

Alex73 [517]3 years ago

6 0

Answer: the answer is 6CO2 + 6H2O C6H12O6 + 6O2

Explanation: I just finished the quiz

!HOPE IT HELPS!

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Describe how water molecules can hydrate various substances

Georgia [21]

Answer:

A water molecule can react with the carbonyl group of an aldehyde or a ketone to form a substance known as a carbonyl hydrate, as shown in the first reaction below. The carbonyl hydrates usually form a very small percentage of the molecules in a sample of a specific aldehyde or ketone.

(Nice profile pic also UwU)

5 0

3 years ago

The compound diborane can be used as a rocket propellant. What is the percent composition of boron in diborane (B2H6)? A. 22% B.

sveticcg [70]

Answer: E. 78%

Explanation:

Diborane (B2H6), a chemical compound made up of approximately 78% Boron and 22% Hydrogen.

4 0

3 years ago

At constant temperature, if a gas occupies 312 mL at a pressure of 1.60 atm, what pressure is necessary for this gas to occupy a

inn [45]

Answer:

P₂ = 1.0 atm

Explanation:

Boyles Law problem => P ∝ 1/V at constant temperature (T).

Empirical equation

P ∝ 1/V => P = k(1/V) => k = P·V => for comparing two different case conditions, k₁ = k₂ => P₁V₁ = P₂V₂

Given

P₁ = 1.6 atm

V₁ = 312 ml

P₂ = ?

V₂ = 500 ml

P₁V₁ = P₂V₂ => P₂ = P₁V₁/V₂ =1.6 atm x 312 ml / 500ml = 1.0 atm

5 0

3 years ago

A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express

Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

$wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63$

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

$V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w$

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

$\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3$

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

$Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3$

We can now calculate the molarity as

$Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}$

8 0

3 years ago

What are the oxidizing agents, and the reducing agents for 2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g)

kondor19780726 [428]

1.) 2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g),

Na is oxidizing agent and H is reducing agent.

2.) C(s)+O2(g)→CO2(g)

C is an oxidizing agent

3.) 2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)

Mn is reducing agent and S is oxidizing agent.

4 0

3 years ago

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