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icang [17]
2 years ago
14

Since the density of air is less than the density of water what will happen to the air if you take a jar of air under water and

Chemistry
1 answer:
nydimaria [60]2 years ago
7 0

Answer:

the air will escape from the jar

Explanation:

Due to its low density air in the jar will be displaced and be replaced by water.

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What other missions had been to Mars, and what did they accomplish?
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A technician needs 500.0mL of a 0.500 M MgCl2 solution for some lab procedures. The stock bottle is labeled 4.00 M MgCl2. What v
mihalych1998 [28]

Answer:

Explanation:

To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;

    Number of moles  = molarity x volume

Parameters given;

        volume of solution = 500mL  = 0.5L

          molarity of solution = 0.5M

   Number of moles  =  0.5 x 0.5  = 0.25moles

Now to know the volume stock to take;

           Volume of stock = \frac{number of moles }{molarity}

  molarity of stock  = 4M

                 volume  = \frac{0.25}{4}   = 0.0625L or 62.5mL

4 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
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