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3 years ago

Studying and got stuck if someone can help me.

Chemistry

1 answer:

Lostsunrise [7]3 years ago

3 0

Answer:

Ph level 0-7 is a acid

ph level 7-14 is base

bases turn paper blue

acids dissolve metal

acids contain hydrogen ions

Explanation:

I dont know the last one all I know that they become neutral when acids and bases are mixed.

You might be interested in

How many grams of KCl is needed to make .75 L of a 1 M solution of KCl?

Zielflug [23.3K]

Answer:

The answer to your question is 25.9 g of KCl

Explanation:

Data

Grams of KCl = ?

Volume = 0.75 l

Molarity = 1 M

Formula

$Molarity = \frac{number of moles}{volume}$

Solve for number of moles

$Number of moles = Molarity x volume$

Substitution

Number of moles = 1 x 0.75

Simplification

Number of moles = 0.75 moles

Molecular mass KCl = 39 + 35.5 = 34.5

Use proportions to find the grams of KCl

34.5 g of KCl ---------------- 1 mol

x ---------------- 0.75 moles

x = (0.75 x 34.5) / 1

x = 25.9 g of KCl

3 0

3 years ago

How would you balance this equation HCL+ Ba > BaCL2 + H2

vladimir2022 [97]

2 HCl + Ba(OH)2 = BaCl2 + 2 H2O is the answer.

5 0

3 years ago

Read 2 more answers

How do the properties of elements change as you move across a period

denpristay [2]

The number of shells increase, the energy level their outer shell have decreases as we go down wards,

3 0

3 years ago

FOR REAL HELPPPPP!!!!

IgorLugansk [536]

Answer:

I belive the answer is gases, liquids, and solids.

Explanation:

The reason I say this is because a vacuum is an empty void of space and time, nothing is there so I believe no sound should be able to penetrate through it.

6 0

3 years ago

If a 32.4 gram sample of sodium sulfate (Na2SO4) reacts with a 65.3 gram sample of barium chloride (BaCl2) according to the reac

STALIN [3.7K]

Answer: The theoretical yield of barium sulfate is 50.9 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For sodium sulfate:

Given mass of sodium sulfate = 32.4 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{32.4g}{142g/mol}=0.228mol$

For barium chloride:

Given mass of barium chloride = 65.3 g

Molar mass of barium chloride = 208.23 g/mol

Putting values in equation 1, we get:

$\text{Moles of barium chloride}=\frac{65.3g}{208.23g/mol}=0.314mol$

The chemical equation for the reaction of barium chloride and sodium sulfate follows:

$Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl$

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of barium chloride

So, 0.228 moles of sodium sulfate will react with = $\frac{1}{1}\times 0.228=0.228mol$ of barium chloride

As, given amount of barium chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate produces 1 mole of barium sulfate.

So, 0.228 moles of sodium sulfate will produce = $\frac{1}{1}\times 0.228=0.228moles$ of barium sulfate

Now, calculating the mass of barium sulfate from equation 1, we get:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.228 moles

Putting values in equation 1, we get:

$0.228mol=\frac{\text{Mass of barium sulfate}}{223.4g/mol}\\\\\text{Mass of barium sulfate}=(0.228mol\times 223.4g/mol)=50.9g$

Hence, the theoretical yield of barium sulfate is 50.9 grams

7 0

3 years ago