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LekaFEV [45]
2 years ago
12

3 common uses for salt

Chemistry
1 answer:
icang [17]2 years ago
7 0

Answer:

Salt has long been used for flavoring and for preserving food. It has also been used in tanning, dyeing and bleaching, and the production of pottery, soap, and chlorine. Today, it is widely used in the chemical industry.

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The phase of matter can be changed by:
stiks02 [169]

Answer: A

Explanation: Mass and volume is the amount, amount doesn’t make the phase of a matter change.

3 0
3 years ago
Read 2 more answers
At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are p
dedylja [7]

Explanation:

The given reaction is as follows.

       2SO_{2} + O_{2}(g) \rightarrow 2SO_{3}(g)

Value of equilibrium constant is given as K_{c} = 4.3 \times 10^{6}[/tex].

Concentration of given species is [SO_2] = 0.010 M; [SO_3] = 10.M; [O_2] = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

             Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

Putting the given concentration as follows.

              Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

             Q = \frac{(10)^{2}}{(0.010)^{2}(0.010)}

              Q = 10^{8}

It is known that when Q > K_{eq}, then reaction moves in the backward direction.

When Q < K_{eq}, then reaction moves in the forward direction.

When Q = K_{eq}, then reaction is at equilibrium.

As, for the given reaction Q > K_{eq} then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.

5 0
3 years ago
A sample of 250 g of water are heated from 40°C to 121°C, calculate the amount of heat energy absorbed.
Marysya12 [62]

Answer:

the anwser isn't in the choices

Explanation:

H=MC(change of temp.)

M=mass of water=250g

C=specific heat of water = 4.186 j/g

change in temperature is 121-40= 81

H= 250x4.186x81=84766.5J

7 0
3 years ago
A field worker is exposed to a xylene for a duration of 8 weeks at 40 hrs/wk. The concentration of xylene in the workplace is 40
Andrej [43]

Answer:

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

Explanation:

Number of hours worker exposed to xylene = 40 hr/week\times 8 week = 320 hours

The concentration of xylene in the workplace =40 \mu g/m^3

The worker is inhaling air at a rate of 0.9 m^3/hr.

Amount xylene inhaled by worker in an hour :

= 40\mu g/m^3\times 0.9 m^3/hr=36 \mu g/hr

Amount xylene inhaled by worker in 320 hours:

36 \mu g/hr\times 320 hr=11,520 \mu g=11,520\times 0.001 mg=11.520 mg

1 μg = 0.001 mg

Amount xylene inhaled by worker in 320 hours = 11.520 mg

1 day = 24 hours

Amount xylene inhaled by worker in 1 day:

\frac{24}{320}\times 11.520 mg=0.864 mg

Assuming 70 kg body mass, the chronic daily intake of xylene :

\frac{0.864 mg/day}{70 kg}=0.01234 mg/ kg day\approx 0.012 mg/ kg day

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

5 0
3 years ago
A solution is made by adding 29.1 mL of concentrated perchloric acid ( 70.5 wt% , density 1.67 g/mL ) to some water in a volumet
Lera25 [3.4K]

Answer:

The concentration of the solution is 1.364 molar.

Explanation:

Volume of perchloric acid = 29.1 mL

Mass of the solution = m

Density of the solution = 1.67 g/mL

m=1.67 g/mL\times 29.1 mL=48.597 g

Percentage of perchloric acid in 48.597 solution :70.5 %

Mass of perchloric acid in 48.597 solution :

= \frac{70.5}{100}\times 48.597 = 34.261 g

Moles of perchloric acid = \frac{34.261 g}{100.46 g/mol}=0.3410 mol

In 29.1 mL of solution water is added and volume was changed to 250 mL.

So, volume of the final solution = 250 mL = 0.250 L (1 mL = 0.001 L)

Molarity=\frac{Moles}{Volume (L)}

=\frac{0.3410 mol}{0.250 L}=1.364 M

The concentration of the solution is 1.364 molar.

6 0
3 years ago
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