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disa [49]
3 years ago
5

How much heat is evolved in converting 1.00 mol of steam at 135.0 ∘c to ice at -45.0 ∘c? the heat capacity of steam is 2.01 j/(g

⋅∘c) and of ice is 2.09 j/(g⋅∘c)?
Chemistry
1 answer:
Viktor [21]3 years ago
4 0
1 mole of steam (H2O) = 18g
Heat evolved = MCФ where m is the mass and c s specific heat capacity while Ф is change in temperature. Latent heat of fusion is 334 kj/kg while latent heat of vaporization is 2260 kj/kg and specific heat of water is 4.2 j/g/c
= 18 ×2.01 × (135-100) = 1266.3 J
    0.018 × 334000 =  6012 J (change of state from gas (steam) to liquid (water)
 18 × 4.186× (100 -0) =  7534.8 J
 0.018 × 2260000 = 40680 J (change of state from liquid to solid ice)
 18 × 2.09 × (0--45) = 1692.9 J
The total heat evolved is therefore 57186 J or 57.186 kJ
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What is observed when magnesium is reacted with hydrochloric acid? choose all that apply
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It is observed that when magnesium reacts with hydrochloric acid, it produces visible bubbles of hydrogen gas.

A vigorous reaction will occur giving rise of heat as it is an exothermic reaction. If heat is applied then it should be more vigorous.

Reaction is as follows:

Mg(s) + HCl (aq) --> MgCl2 (aq) + H2(g)

Magnesium reacts easily with HCl to produce H2 gas and magnesium ions, Mg2+, and heat. The reaction is exothermic, so it heats up quickly.

Mg(s) + HCl (aq) --> MgCl2(aq) + H2(g)

The net ionic equation :

Mg(s) + 2H+ --> Mg2 + H2(g)

If water is removed from the solution then white crystals of Mgcl2 is obtained.

Or in simple words,

2Mg + 2HCl -> 2Mg+ + 2Cl- + H2(gas)

The magnesium is attacked by the hydrochloric acid resulting in the magnesium dissolving into the solution resulting in a solution of magnesium chloride in hydrochloric acid and the production of hydrogen gas.

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3 0
1 year ago
When a 5.00-g metal piece, A, was immersed in 38.0 mL of water, the water level rose to 50.0 mL. Similarly, when a 5.00-g metal
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Answer:

A is denser than B as it's volume for the same mass is smaller.

Explanation:

Hello.

In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:

V_A=50.0mL-38.0mL=12.0mL\\\\V_B=60.0mL-38.0mL=22.0mL

Next, we compute the densities as shown below:

\rho_A=\frac{m_A}{V_A}=\frac{5.00g}{12.0mL}=0.42g/mL\\  \\\rho_B=\frac{m_B}{V_B}=\frac{5.00g}{22.0mL}=0.23g/mL

In such a way, A is denser is B as it's volume for the same mass is smaller.

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