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3 years ago

What does it mean to control variables in an experiment?

Chemistry

2 answers:

Goshia [24]3 years ago

7 0

D) to keep as many things the same as possible across the parts of an experiment.

Anit [1.1K]3 years ago

4 0

Answer:

Explanation:

I assume its D because the controlled variable is what is kept the same throughout the experiment.

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Molarity s defined as the

RideAnS [48]

The molar concentration of a solution, usually expressed as the number of moles of solute per liter of solution.

3 0

3 years ago

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Write the balanced chemical equation for the dissociation reaction of thiosulfuric acid.

mafiozo [28]

Answer:

f(x)=3x^2

Explanation:

5 0

3 years ago

G8_SCIENCE_BSA_21_22

S_A_V [24]

Answer:

Explanation: Over the course of a year, areas near Earth’s equator receive more direct energy from the Sun than areas near Earth’s poles. Which of the following correctly describes one result of this heating pattern?

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3 years ago

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa

notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$

Finally, the percent yield turns out into:

$Y=\frac{1.92g}{7.1g} *100$

$Y=27.0$ %

Best regards.

6 0

3 years ago

A 49.48-mL sample of an ammonia solution is analyzed by titration with HCl. It took 38.73 mL of 0.0952 M HCl to titrate the ammo

Kryger [21]

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

8 0

3 years ago