CHEESE TOES ARE KING

There is a close relation between forest<br>and wildlife.

evablogger [386]

Answer:

habitat

Explanation:

wild animal need

4 0

3 years ago

Which of the mole ratios would be used to set up the below problem?

san4es73 [151]

A.3 moles H2

1 mole N2

I think it is correct

4 0

3 years ago

Ammonia, NH3NH3 , can react with oxygen to form nitrogen gas and water. 4NH3(aq)+3O2(g)⟶2N2(g)+6H2O(l) 4NH3(aq)+3O2(g)⟶2N2(g)+6H

solmaris [256]

Answer:

36.37% is the percent yield of the reaction.

Explanation:

$4NH_3(aq)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)$

1)0.650 L nitrogen gas , at 295 K and 1.01 bar.

Let the moles of nitrogen gas be n.

Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)

Temperature of the gas = T = 295 K

Volume of the gas = V = 0.650 L

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9967 atm\times 0.650 L}{0.0821 atm L/mol K\times 295 K}=0.0267 mol$

2) Moles of ammonia gas= $\frac{2.53 g}{17 g/mol}=0.1488 mol$

Moles of oxygen gas = $\frac{3.53 g}{32 g/mol}=0.1101 mol$

According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.

Then,0.1101 mol of oxygen will react with:

$\frac{4}{3}\times 0.1101 mol=0.1468 mol$ of ammonia.

Hence, oxygen gas is in limiting amount and act as limiting reagent.

3) Theoretical yield of nitrogen gas :

According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.

Then 0.1101 mol of oxygen will give:

$\frac{2}{3}\times 0.1101 mol=0.0734 mol$ of nitrogen.

Theoretical yield of nitrogen gas = 0.0734 mol

Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol

Percentage yield:

$\frac{\text{Experiential yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of the reaction:

$\frac{ 0.0267 mol}{0.0734 mol}\times 100=36.37\%$

36.37% is the percent yield of the reaction.

3 0

3 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

2

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

So

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$

The rate constant is $0.4167\ \text{M}^{-1}\text{min}^{-1}$

3 0

3 years ago

You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that t

jek_recluse [69]

Answer: D. Mutation in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

Explanation: It was not likely to be that the coding sequences are replicated more often. The only possible explanation is that the mutations in coding is more likely to be deleterious to the organism than mutations because it is in a non coding sequence.

8 0

3 years ago