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3 years ago

One mole of copper(ii) sulfate, cuso4, contains ________ o atoms.

Chemistry

2 answers:

Naddik [55]3 years ago

8 0

Answer : One mole of copper(II)sulfate contains $24.088\times 10^{23}$ number of oxygen atoms.

Solution :

As we know that, 1 mole of contains $6.022\times 10^{23}$ number of atoms.

In $CuSO_4$ , there are 1 number of copper atom, 1 number of sulfur atom and 4 number of oxygen atoms.

As 1 mole of copper(II)sulfate contains $4\times 6.022\times 10^{23}=24.088\times 10^{23}$ number of oxygen atoms.

Therefore, one mole of copper(II)sulfate contains $24.088\times 10^{23}$ number of oxygen atoms.

Ierofanga [76]3 years ago

6 0

According to avogadro constant, the number of units in one mole of any substance contain 6.022 x10 ^23 atoms

therefore the number of o atoms in one mole of CuSO4 = 6.022 x 10 ^ 23

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A 45ml of a 4M solution of CaBr2 contains how many grams of CaBr2?

natta225 [31]

The required mass of calcium bromide is 35.98 grams.

<h3>What is molarity?</h3>

Molarity is any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V, where

M = molarity = 4M
V = volume = 45mL = 0.045L

Moles will be calculated by using the above equation as:

n = (4)(0.045) = 0.18 mole

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

W = given mass
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Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g

Hence required mass of CaBr₂ is 35.98 grams.

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2 years ago

Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid w

Naddika [18.5K]

Answer:

87.9 % is the percent yield of H₂O

Explanation:

This is the neutralization reaction. A base reacts with an acid to produce water and the correspondly ionic salt.

NaOH + HCl → NaCl + H₂O

As we have the mass of the two reactants, we must determine the limiting reactant.

Let's convert to moles, the mass of each reactant. (mass / molar mass)

21.1 g / 36.45 g/mol = 0.579 moles of HCl

46.3 g / 40g/mol = 1.15 moles of NaOH

Ratio is 1:1, so it is obviously that the limiting reactant is the HCl. For 1.15 moles of NaOH, i need the same amount of acid, but I only have 0.579 moles

Let's work with the products now. Ratio is 1:1 again, so If I have 0.579 moles of acid, I can produce 0.579 moles of H₂O.

How many grams are 0.579 moles of water? We should find it out as this

mol . molar mass = mass → 0.579 mol . 18 g/mol = 10.4 g

We were told that the production of water was 9.17 g, so let's determine the percent yield as this:

(Yield produced / Theoretical yield) . 100 =

(9.17 g / 10.4g ) . 100 = 87.9 %

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3 years ago

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3 years ago

Chemists use the periodic table to organize the elements. What are some ways in which this organization is useful? Select all th

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2 years ago

Read 2 more answers

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76

aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.

The no. of moles can be determined by using the formula,

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.

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3 years ago