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3 years ago

Refusing to take part in sexual activity is called sexual abstinence? True or false.

Chemistry

2 answers:

Temka [501]3 years ago

5 0

Answer:

true.

Explanation:

Diano4ka-milaya [45]3 years ago

3 0

Answer:

yes because you can look it up??

Explanation:

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Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.

kiruha [24]

Answer:

Explanation:

For pH of a buffer solution , the formula is

pH = pKa + log [ Base ] / [ conjugate acid ]

= pKa + log [ NH₃ ] / [ NH₄⁺ ]

Ka = Kw / Kb

Kb for NH₄OH = 1.8 x 10⁻⁵

Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵

= 5.6 x 10⁻¹⁰

pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2

= 10 - log 5.6

= 9.25

Effect of addition of HCl

H⁺ of HCl will react with NH₃ to produce NH₄⁺

25 mL of .1 HCl = 2.5 mM of HCl

25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺

65 mL of .2 M NH₃ = 13 mM of NH₃

65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺

NH₃ + H⁺ = NH₄⁺

NH₄⁺ formed = 2.5 + 13 mM

15.5 mM of NH₄⁺

NH₃ = 13 mM

Concentration of NH₃ = 13 / 90

Concentration of NH₄⁺ = 15.5 / 90

pH of final buffer mixture

= 9.6 + log 13 / 15.5

= 9.25 - .076

= 9.174

3 0

4 years ago

Does changing the number of neutrons affect the atomic number?

mixer [17]

Answer:

awsdxcvawsd

Explanation:

5 0

4 years ago

Read 2 more answers

Clouds can form in all the following ways except

stich3 [128]

What are the options?

4 0

3 years ago

At 500°C the equilibrium constant, Kp, is 4.00 × 10–4 for the equilibrium: 2HCN(g) H2(g) + C2N2(g) What is Kp for the following

Andrews [41]

<u>Answer:</u> The value of $K_p$ for the equation is $2.5\times 10^3$

<u>Explanation:</u>

The given chemical reaction is:

$2HCN(g)\rightleftharpoons H_2(g)+C_2N_2(g);K_1$

The chemical equation for which the equilibrium constant is to be calculated follows:

$H_2(g)+C_2N_2(g)\rightleftharpoons 2HCN(g);K_p$

As, the equation is the result of the reverse of given reaction. So, the equilibrium constant for the equation will be the inverse of equilibrium constant for the given reaction.

The value of equilibrium constant for the equation is:

$K_p=\frac{1}{K_1}$

We are given:

$K_1=4.00\times 10^{-4}$

Putting values in above equation, we get:

$K_p=\frac{1}{4.00\times 10^{-4}}=2.5\times 10^3$

Hence, the value of $K_p$ for the equation is $2.5\times 10^3$

3 0

4 years ago

If a solution of glucose has a concentration of 0.500 M, how many moles of are in 8.00 L of

Sidana [21]

Answer:

n = 4mol

Explanation:

n =? C= 0.5M, V = 8L

Substitute into

n = CV

n= 0.5×8= 4mol

6 0

3 years ago