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Alexus [3.1K]
3 years ago
9

The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Chemistry
1 answer:
Alona [7]3 years ago
3 0

Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

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Explanation:

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6 0
3 years ago
how many moles of sodium hydroxide would have to be added to 250 ml of a 0.403 m acetic acid solution, in order to prepare a buf
jonny [76]

The volume of the buffer solution having a ph value is calculated by henderson's hasselbalch equation.

Buffer solution is water based solution which consists of a mixture containing a weak acid and a conjugate base of the weak acid. or a weak base and conjugate acid of a weak base.it is a mixture of weak acid and a base. The pH of the buffer solution is determined by the expression of the henderson hasselbalch equation.

              pH=pKa + log [salt]/[acid]

Where, pKa =dissociation constant , A- = concentration of the conjugate base, [HA]= concentration of the acid. Here, a buffer solution contains 0.403m acetic acid  and 250 ml is added  in order to prepare a buffer with a ph of 4.750. Putting all the values in the henderson hasselbalch equation we find the pH of the buffer solution.

To learn more about hendersons hasselbalch equation please visit:

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6 0
1 year ago
A sample contains 16 mg of polonium-218. After 12 minutes, the sample will contain 1.0 mg of polonium-218. What is the half life
mariarad [96]

Answer:

Half-life = 3 minutes

Explanation:

Using the radioactive decay equation we can solve for reaction constant, k. And by using:

K = ln2 / Half-life

We can find half-life of polonium-218

Radioactive decay:

Ln[A] = -kt + ln [A]₀

Where:

[A] could be taken as mass of polonium after t time: 1.0mg

k is Reaction constant, our incognite

t are 12 min

[A]₀ initial amount of polonium-218: 16mg

Ln[A] = -kt + ln [A]₀

Ln[1.0mg] = -k*12min + ln [16mg]

-2.7726 = - k*12min

k = 0.231min⁻¹

Half-life = ln 2 / 0.231min⁻¹

<h3>Half-life = 3 minutes</h3>

5 0
3 years ago
What fuel source is Jan using if she exercises at 85% of her maximum aerobic capacity?
Marina86 [1]

Answer:

Carbohydrates

Explanation:

Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.

In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.

4 0
3 years ago
The density of water at 30.0 °C is 0.9956 g/mL. If the specific gravity of acetic acid is 1.040 at 30.0 °C, what is the density
mash [69]

Answer:

The density of acetic acid at 30°C = 1.0354_g/mL

Explanation:

specific gravity of acetic acid = (Density of acetic acid at 30°C) ÷ (Density of water at 30°C)

Therefore, the density of acetic acid at 30°C = (Density of water at 30°C) × (Specific gravity of acetic acid at 30°C)

= 0.9956 g/mL × 1.040

= 1.0354_g/mL

Specific gravity, which is also known as relative density, is the ratio of the density of a substance to the density of a specified standard substance.

Generally the standard substance of to which other solid and liquid substances are compared is water which has a density of 1.0 kg per litre or 62.4 pounds/cubic foot at 4 °C (39.2 °F) while gases are normally compared with dry air, with a density of 1.29 grams/litre or 1.29 ounces/cubic foot under standard conditions of a temperature of 0 °C and one standard atmospheric pressure

7 0
3 years ago
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