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nignag [31]
3 years ago
14

Why is salt put on icy roads and sidewalks in the winter?

Chemistry
1 answer:
balandron [24]3 years ago
3 0
So that ice would not melts and it helps to keep grip for walking or driving
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What are some facts about animal cells
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Compared to plant cells, which are shaped somehow like a box, animal cells are round and circular.

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3 years ago
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¿cual es el volumen de solución que contiene 10 gramos de NaCl en 100 mL de solución?.
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3 years ago
Acetylene is the substance burned in oxy-acetylene torches. Write a balanced equation for the complete oxidation reaction that o
nexus9112 [7]

Answer: 2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products must be same as the mass of reactants.

For the conservation of mass, the number of atoms of each element must be same in reactants and products. Thus we need to balance the reaction by writing appropriate stochiometric coefficients.

All the hydrocarbons burn in oxygen to form carbon dioxide and water.Thus the complete balanced equation is:

2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O

4 0
3 years ago
A sample of aluminum has mass 26.98 g and has a density of
kherson [118]

Answer:

9.99L

Explanation:

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8 0
3 years ago
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How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The
bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz =  100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0
4 years ago
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