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Darya [45]
3 years ago
12

What is a proton?

Chemistry
1 answer:
Scrat [10]3 years ago
3 0

Answer:

the positively charged particle found in atomic nuclel

Explanation:

protons carry positive charge and attracts negatively charged particles called electron

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What volume of methanol should be used to make 1.25 l of a 21.0 by volume solution
Nimfa-mama [501]
Answer:
262.5 mL

Explanation:
21% of 1.25 L volume is methanol, so the amount of methanol is,
0.21 × 1250 mL = 262.5 mL
6 0
3 years ago
Phenylacetic acid, HC8H7O2 is one of the substances that accumulates in the blood of persons with phenylketonuria, an inherited
STALIN [3.7K]
<span>Ka = [H+][PAc-] / [PAcH]

[H+] = 10^-2.68 = 0.00209M = [PAc-]
[PAcH] = 0.085 = 0.00209 = 0.083

Ka = (.00209)(.00209) / (0.083) = Answer Here.</span>
3 0
4 years ago
: Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050-L flask at 668 K, how many moles of Cl2(g) will be present
Mrac [35]

Answer:

The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

Explanation:

The reaction is:

CO(g) + Cl₂(g) ⇄ COCl₂(g)  

The equilibrium constant of the above reaction is:

K = 1.2x10³

To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

The equilibrium constant for the reverse reaction is:

K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4}

Now, we need to calculate the concentration of CO and COCl₂:

C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M

C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M

Now, from the reaction we have:

COCl₂(g) ⇄ CO(g) + Cl₂(g)  

0.018 - x       0.115+x   x    

The concentration of Cl₂ is:

K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]}

8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x}  

8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0  

By solving the above equation for x we have:

x = 1.29x10⁻⁴ M = [Cl₂]

Finally, the number of moles of Cl₂ present at equilibrium is:

\eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles

Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.

I hope it helps you!

8 0
3 years ago
Data that is collected that is mostly numbers?
Mekhanik [1.2K]

Answer:

statistics

Explanation:

statistics is the study of using numbers to express data

7 0
3 years ago
Respiration describes the process that living cells use to release energy by combining sugar and oxygen. The primary chemical ch
True [87]

Answer:

C the mass of the reactant molecules is the same as the mass of the product

5 0
3 years ago
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