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3 years ago

I'm giving the brainliest if it's correct thank you

Chemistry

1 answer:

andriy [413]3 years ago

8 0

Answer:

Explanation:

Now, you know that this solution has a molarity of 6.00 M, which basically means that every liter, which is the equivalent of 1000 mL, will contain 6.00 moles of ammonia. Since we've picked a sample of 1000 mL, you can say that it will contain 6.00 moles of ammonia. To convert this to grams, use the compound's molar mass

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Whats the si unit chemist use to measure the amount of an element substance?

Westkost [7]

The SI unit for the amount of substance present is the mole.

The mole is defined as the amount of substance that has the same amount of particles as there are atoms in 12 grams of carbon-12. Mathematically, the moles of a substance may be computed using:

moles present = mass of substance / molecular mass of substance

4 0

3 years ago

ASAP!!!!!!!!!!!!

dlinn [17]

a. Tall prarie grass burns after being struck by lightning.

6 0

2 years ago

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

2 years ago

In the laboratory you dissolve 15.9 g of barium chloride in a volumetric flask and add water to a total volume of 375 ml. what i

ASHA 777 [7]

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. So, we calculate as follows:

Molarity = 15.9 g BaCl2 ( 1 mol / 208.23 g ) / .375 L = 0.204 mol / L

5 0

3 years ago

How many moles of BaSO4 are formed if 0.5 moles of Na2 SO4 react with 60 g of BaCl2? Na2 SO4 + BaCl2 → BaSO4 + 2NaCl

faust18 [17]

the balanced equation for the reaction is as follows
Na₂SO₄ + BaCl₂ ----> 2NaCl + BaSO₄
stoichiometry of Na₂SO₄ to BaCl₂ is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.

number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other

therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl₂ to BaSO₄ is 1:1.

therefore number of BaSO4 moles formed - 0.288 mol

6 0

3 years ago

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