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Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>
Emulsions are an example of colloids<span> composed of tiny particles </span>suspended<span> in another immiscible (unmixable) material. An emulsion is a </span>suspension<span> of two liquids that usually do not mix together. These liquids that do not mix are said to be immiscible. An example would be </span>oil and water<span>.
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<span>0.48 grams.
Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced.
To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So
0.0075 mol * 63.546 g/mol = 0.476595 g.
Round the results to 2 significant figures, giving 0.48 grams.</span>
The heat used in phase changes is calculated by multiplying the mass of the substance by the energy of the phase change. In this case, for liquid to boil, we would find total heat by multiply the mass of liquid by the latent heat of vaporization (Hvap). If we are instead given the Hvap and the total heat of 1 kJ, we would divide 1 kJ by the Hvap (which is usually in kJ/kg) to get the mass of liquid boiled (in kg).
