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3 years ago

What structural units make up network solids?

Chemistry

2 answers:

My name is Ann [436]3 years ago

8 0

Non metal atoms got this from google btw

slega [8]3 years ago

5 0

Answer:

Nonmetal atoms

Explanation:

A P E C C

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A gas occupies 14.3 liters at a pressure of 45.0 mm Hg. What is the volume when the pressure is increased to

Simora [160]

Answer:

P1V1= P2V2

Explanation:

Inverse relationship

V2 = V1 X P1/P2

V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99

8 0

3 years ago

A 20.0 mL solution of NaOH is neutralized with 24.1 mL of 0.200 M HBr. What is the concentration of the original NaOH solution

Alinara [238K]

Answer:

0.241 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HBr + NaOH —> NaBr + H₂O

From the balanced equation above,

The mole ratio of acid, HBr (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 1

Finally, we shall determine the concentration of the NaOH solution. This can be obtained as follow:

Volume of base, NaOH (V₆) = 20 mL

Volume of acid, HBr (Vₐ) = 24.1 mL

Concentration of acid, HBr (Cₐ) = 0.2 M

Concentration of base, NaOH (C₆) =?

CₐVₐ / C₆V₆ = nₐ/n₆

0.2 × 24.1 / C₆ × 20 = 1/1

4.82 / C₆ × 20 = 1

Cross multiply

C₆ × 20 = 4.82

Divide both side by 20

C₆ = 4.82 / 20

C₆ = 0.241 M

Therefore, the concentration of the NaOH solution is 0.241 M

8 0

3 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

Answer:

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

6 0

3 years ago

What volume of carbon dioxide gas can be collected from

alisha [4.7K]

Answer:

1.22 L of carbon dioxide gas

Explanation:

The reaction that takes place is:

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we determine which reactant is limiting:

Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃

Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

0.05 mol HCl * $\frac{1molCO_{2}}{1molHCl}$ = 0.05 mol CO₂

Finally we use PV=nRT to calculate the volume:

P = 1 atm

n = 0.05 mol

T = 25 °C ⇒ 25 + 273.16 = 298.16 K

1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

V = 1.22 L

7 0

3 years ago

Does the tempurture of the vinegar affect the volcano flow

aleksandr82 [10.1K]

The temperature of vinegar does increase the rate of reaction according to the collision theory.

Hope this answers your question!

6 0

2 years ago