The number of protons + number of neutrons is equal to the

What is the purpose of cell respiration? to breath in food to expel unwanted oxygen to produce energy from food]

zysi [14]

Answer:

cellular respiration is a critical function by which cells release energy for various cellular activities like locomotion, biosynthesis, and even the transportation of molecules between membranes.

3 0

3 years ago

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows:

ankoles [38]

Answer:

The answer to your question is: 68.3%

Explanation:

Urea = ?

Theoretical yield

Ammonia 145.1 kg

Carbon dioxide = 211.4 kg

Urea = 174.9 kg

MW NH3 = 17kg

MW CO2 44 kg

MW Urea = 60 kg

2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l)

34 44 kg 60kg

145.1 211.4 174.9

Limiting reactant = NH3

34kg of NH3 ------------------ 60 kg of Urea

145.1 kg ------------------- x

x = (145.1 x 60) / 34

x = 256 kg of Urea

Percent yield = 174.9 / 256 x 100

= 68.3%

7 0

3 years ago

PLZZZZ HELP MEH THANKS YEET

aleksandrvk [35]

Remark

The question with these kind of problems is "Which R do you use?" That's where dimensional analysis is so handy. You must look at the units of the givens and choose your R accordingly. You'll see how that works in a moment.

You need to list the givens along with their units and in this case the property you want to solve for. You need all that to determine the R value

Givens

n = 0.25 moles

T = 35°C = 35 + 273.15 = 308.15°K

V = 6.23 L

Pressure = P in kPa

Which R

The units of the R you want has to have units of moles, kPa, °K and liters

The R that you want is 8.314

Formula

PV = nRT

P 6.23 = 0.25 * 8.314 * 308.15 Combine the left

P*6.23 = 640.5

P = 640.5/6.23 = 102.81 The answer should be 100 kpA of 1.0 * 10^2 kPa

because the number of moles has only 2 sig digs.

But if sig digs are not a problem 102.8 is likely close enough.

Second Question

You are going to have to clean up the numbers. I think I've got only 1 chance at this. The partial pressures of the 2 gases will add up to the total pressure. So the total pressure was 100 approx and the water vapor was 3.36 kPa. The difference is

Total = air + water vapor

100.18 = air + 3.36 about Subtract 3.36 from both sides.

100.18 - 3.36 = 96.82 about. Pick the answer that is closest to that. I'll clean up the numbers if I can.

Answer C

6 0

3 years ago

Read 2 more answers

A student made a graph to show the chemical equilibrium position of a reaction. The student forgot to label the y-axis of the gr

Arturiano [62]

You have to get the x said

3 0

2 years ago

Read 2 more answers

Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.65 g of N2H4 reacts with

White raven [17]

Answer: The percent yield of the reaction is 17.41 %.

Explanation:

For $N_2H_4$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $N_2H_4$ = 2.65 g

Molar mass of $N_2H_4$ = 32.04 g/mol

Putting values in above equation, we get:

$\text{Moles of }N_2H_4=\frac{2.65g}{32.04g/mol}=0.0827mol$

For $N_2$

To calculate the number of moles, we use the equation given by ideal gas equation:

$PV=nRT$

where,

P = pressure of the gas = 1 atm

V = Volume of gas = 0.350 L

n = Number of moles = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = 273 K

Putting values in above equation, we get:

$1.00atm\times 0.350L=n\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\n=0.0144mol$

Now, to calculate the experimental yield of $N_2$ , we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

We are given:

Moles of nitrogen gas = 0.0144 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$0.0144mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=0.4032g$

Experimental yield of nitrogen gas = 0.4032 g

For the given chemical equation:

$N_2H_4(aq.)+O_2(g)\rightarrow N_2(g)+2H_2O(l)$

By Stoichiometry of the reaction:

1 mole of $N_2H_4$ produces 1 mole of nitrogen gas.

So, 0.0827 moles of $N_2H_4$ will produce = $\frac{1}{1}\times 0.0827=0.0827mol$ of nitrogen gas.

Now, to calculate the theoretical yield of nitrogen gas, we use equation 1:

Moles of nitrogen gas = 0.0827 mol

Molar mass nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$0.0827mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=2.3156g$

Theoretical yield of nitrogen gas = 2.3156 g

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 0.4032 g

Theoretical yield of nitrogen gas = 2.3156 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{0.4032g}{2.3156g}\times 100\\\\\% \text{yield of nitrogen gas}=17.41\%$

Hence, the percent yield of the reaction is 17.41 %.

8 0

3 years ago