The properties of a substance are not affected by chemical reactions. O True O False

Beaker A contains 2.06 mol of copper ,and Barker B contains 222 grams of silver.Which beaker the larger number of atom?

Dmitry [639]

Answer:

The number of copper atoms 12.405 ×10²³ atoms.

The number of silver atoms 13.13 ×10²³ atoms.

Beaker B have large number of atoms.

Explanation:

Given data:

In beaker A

Number of moles of copper = 2.06 mol

Number of atoms of copper = ?

In beaker B

Mass of silver = 222 g

Number of atoms of silver = ?

Solution:

For beaker A.

we will solve this problem by using Avogadro number.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.

While we have to find the copper atoms in 2.06 moles.

So,

63.546 g = 1 mole = 6.022×10²³ atoms

For 2.06 moles.

2.06 × 6.022×10²³ atoms

The number of copper atoms 12.405 ×10²³ atoms.

For beaker B:

107.87 g = 1 mole = 6.022×10²³ atoms

For 222 g

222 g / 101.87 g/mol = 2.18 moles

2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms

8 0

3 years ago

What is the mass of OH- in Ca(OH)2

Ne4ueva [31]

Oxygen: 16.00 g/mol
Hydrogen: 1.01 g/mol

The mass would be 17.01 g/mol

5 0

3 years ago

A 0.0000792 M sample of Compound X in a solvent has an absorbance of 0.341 at 528 nm in a 1.000-cm cuvet. The solvent alone has

natima [27]

Answer:

a) a = 3485 M⁻¹cm⁻¹

b) C = 0,000127 M

Explanation:

Lambert-Beer law says that there is a linear relationship between absorbance and concentration of a chemical substance. The formula is:

A = a×b×C

Where A is absorbance, a is molar absorptivity, b is path length and C is concentration.

a) In the problem Concentration is 0.0000792 M, b is 1,000cm and Absorbance is absorbance of sample-absorbance of blank: 0,341-0,065 = 0,276

Replacing:

0,276 = a×1,000cm×0,0000792M

a = 3485 M⁻¹cm⁻¹

b) As the experiment consist in the same compound in the same solvent, the molar absorptivity will be the same, a = 3485 M⁻¹cm⁻¹, path length will be 1,000cm and absorbance: 0,508-0,065 = 0,443

Replacing:

0,443 = 3485 M⁻¹cm⁻¹×1,000cm×C

C = 0,000127 M

I hope it helps!

8 0

2 years ago

Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma

Elena-2011 [213]

Answer:

The yearly release of $CO_2$ into the atmosphere is $6.73\times 10^{10} kg$ .

Explanation:

$CaCO_3(s)\rightarrow CaO(s) + CO_2(g)$

Annual production of CaO = $8.6\times 10^{10} kg=8.6\times 10^{13} g$

Moles of CaO :

$\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles$

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with $1.53\times 10^{12} moles$ of CaO moles of carbon-dioxide moles produced will be:

$\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles$ of carbon-dioxide

Mass of $1.53\times 10^{12}$ moles of carbon-dioxide:

$1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg$

The yearly release of $CO_2$ into the atmosphere is $6.73\times 10^{10} kg$ .

6 0

3 years ago

How do plants most commonly break large rocks into smaller pieces

galina1969 [7]

Plants most commonly break large rock into smaller pieces by having the plant root grow into cracks in rocks. The plant root from below the surface grows and that's how they break rocks into pieces.

4 0

3 years ago

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