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IgorLugansk [536]

3 years ago

8

In preparation for a blizzard, a city dispatches a crew to spread salt on the street surface of the bridges around town. The sal

t is spread on the streets because it will likely change the freezing point of water from ____ to ____.

1 answer:

timofeeve [1]3 years ago

4 0

Solid to liquid because it wil defrost

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Two forces that are not equel in size are

satela [25.4K]

Balanced and Unbalanced Forces

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3 years ago

What does a plant produce during photosynthesis? A. Oxygen and glucose B. Carbohydrates and carbon C. Carbon, oxygen, and hydrog

Nataliya [291]

Answer:

im pretty sure b

Explanation:

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4 years ago

A force that acts on rock to change its shape or volume is called

sineoko [7]

Answer:

The answer would be stress

Explanation:

A force that acts on rock to change its shape or volume is stress. Stress adds energy to the rock. The energy is stored in the rock until it changes shape or breaks. Three different kinds of stress can occur in the crust—tension, compression, and shearing.

I hope this helps

8 0

3 years ago

Identify solute and solvent in 80 solution of ethyl alcohol with water

diamong [38]

The solute is the part of the solution that dissolves in the second component (usually a fluid). Therefore, for the mentioned solution, the solute is ehyl alcohol since it is the one dissolving in water.

As for the solvent, it is the component in which the solute dissolves. In this case, it is water.

5 0

3 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

3 0

3 years ago