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4 years ago

How many formula units make up 12.8 g of magnesium chloride (MgCl2)? Express the number of formula units numerically.

Chemistry

1 answer:

Cloud [144]4 years ago

6 0

<h3>Answer:</h3>

8.09 × 10²² formula units

<h3>Explanation:</h3>

Mass of MgCl₂ is 12.8 g

Required to calculate the formula units

1 mole of a compound contains the Avogadro's number formula units

That is;

1 mole = 6.022 × 10²³ formula units

<h3>Step 1: Calculate the moles of MgCl₂ in 12.8 g </h3>

The molar mass of MgCl₂ is 95.211 g/mol

Number of moles = Mass ÷ molar mass

Moles of MgCl₂ = 12.8 g ÷ 95.211 g/mol

= 0.1344 mole

<h3>Step 2: Calculate the formula units of MgCl₂</h3>

1 mole = 6.022 × 10²³ formula units

Therefore;

Number of formula units = Moles × 6.022 × 10²³ formula units

= 0.1344 mole × 6.022 × 10²³ formula units

= 8.09 × 10²² formula units

Therefore, there are 8.09 × 10²² formula units in 12.8 g of MgCl₂

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The products of a reaction are NaCl and H2O. What does the law of conservation of matter reveal about the reactants in the react

allochka39001 [22]

Answer:

Explanation:

total mass should be equal on both sides

it's a double do displacement reaction

HCl + NaOH --> NaCl + H2O

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3 years ago

Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.

Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

The mole ratio of the acid, H₃PO₄ (nA) = 2
The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of acid, H₃PO₄ (Ma) = 0.390 M
Volume of acid, H₃PO₄ (Va) = 24.5 mL
Molarity of base, Ca(OH)₂ (Mb) = 0.279 M

Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

Learn more about titration:

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2 years ago

When you add salt to water it makes the solution boil

alukav5142 [94]

Answer:

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Explanation:

4 0

3 years ago

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Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made

arlik [135]

Answer: The empirical formula is $C_3H_3O$ .

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 12.24 g

Mass of $H_2O$ = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, = $\frac{12}{44}\times 12.24=3.338g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, = $\frac{2}{18}\times 2.505=0.278g$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.278}{0.104}=3$

For H = $\frac{0.278}{0.104}=3$

For O = $\frac{0.104}{0.104}=1$

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is $C_3H_3O$ .

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3 years ago

Write a ground state electron configuration for each neutral atom

Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0

3 years ago