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3 years ago

Please help me and please be correct

Chemistry

1 answer:

Alex3 years ago

4 0

I will go for the last one

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17) What is the molarity

chubhunter [2.5K]

Heya it seems you’ve made a mistake in one of your plausible answers.. as for me l got 0.64M so l am thinking the option a will be in a decimal form

6 0

3 years ago

A sample of gas is at 273°K and 2.00 atm with a volume of 6.00 liters. When the gas is compressed to 3.00 liters and the pressur

horrorfan [7]

Answer: 0.00488K

Explanation: using the general gas equation

(P1V1)/T1 =(P2V2)/T1

Substitute

2*6/273= 3*3/T

Simplify

T= 0.00488°K

3 0

4 years ago

PV = nRT. If P = 1 atm, V = 5.0 liter, R = 0.0821 L.atm/mol.K, and T = 293 K; what is the value of n?

Leya [2.2K]

Answer:

n = 0.207 mole

Explanation:

We have,

P = 1 atm

V = 5 liter

R = 0.0821 L.atm/mol.K

T = 293 K

We need to find the value of n. The relation is as follows :

PV = nRT

Solving for n,

$n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 5}{0.0821 \times 293}\\\\n=0.207\ \text{mol}$

So, the value of n is 0.207 mol.

6 0

3 years ago

How many moles of nitrogen are in 2.0×10−2mole of quinine?

galben [10]

Answer: 4 x 10 ∧-2 moles of nitrogen.

Explanation:

The chemical formular for quinine is ; C20 H24 N2 O2

As can be seen from the chemical formular;

1 mole of quinine contains 2 moles of Nitrogen

Thus; 2.0 x 10 ∧-2 moles of quinine would contain

2.0 x 10 ∧-2 x 2 = 4 x 10 ∧-2 moles of nitrogen.

Therefore 4 x 10 ∧-2 moles of nitrogen are in 2.0×10−2mole of quinine

7 0

4 years ago

A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the

Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

$\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}$

R is Gas constant having value = 0.008314 kJ / K mol

Given temperature, T = 293 K

Given, $\Delta{G^0}=15\ kJ/mol$

So, Applying in the equation as:-

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

Thus,

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

$\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}$

$\frac{[His]}{[His+]}=0.00211$

Also, considering:-

$pH=pKa+log\frac{[His]}{[His+]}$

Given that:- pH = 7.0

So, $7.0=pKa+log0.00211$

<u>pKa of the histidine = 9.67</u>

8 0

3 years ago