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Vedmedyk [2.9K]
2 years ago
15

If the pH is 10, what is the concentration of hydroxide ion?

Chemistry
1 answer:
AleksandrR [38]2 years ago
7 0

Answer : The correct option is, 10^{-4}M

Explanation : Given,

pH = 10

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10=4

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

4=-\log [OH^-]

[OH^-]=1.0\times 10^{-4}M

Therefore, the OH^- concentration is, 1.0\times 10^{-4}M

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Identify the intermolecular forces present in each of these substances nh3 hcl co co2
nasty-shy [4]
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4 0
3 years ago
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A slice of Swiss cheese contains 47 mg of sodium. (a) What is this mass in grams? .047 (b) What is this mass in ounces? (16 oz =
MAVERICK [17]

Answer:

(a) 0.047 g (b) 0.0016 oz (c) 0.0001 lb

Explanation:

The given mass of the sodium in the slice = 47 mg

(a) Mass has to be calculated in grams

The conversion of mg to g is shown below as:

1 mg = 10⁻³ g

So,

<u>Mass of sodium = 47 × 10⁻³ g = 0.047 g</u>

(b) Mass has to be calculated in ounces

The conversion of ounces to g is shown below as:

453.6 g = 16 oz

Or,

1 g = 16 / 453.6 oz

So,

<u>Mass of sodium = (0.047 × 16) / 453.6 oz = 0.0016 oz</u>

(c) Mass has to be calculated in pounds

The conversion of pounds to g is shown below as:

1 lb = 453.6 g

Or,

1 g = 1/ 453.6 lb

So,

<u>Mass of sodium = (0.047 × 1) / 453.6 oz = 0.0001 lb</u>

5 0
3 years ago
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

Now put all the given values in this formula, we get

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

7 0
3 years ago
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