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Anastaziya [24]
3 years ago
5

Why is it reasonable to assume the specific heats of naoh and hcl solutions are the same as water?

Chemistry
1 answer:
ivolga24 [154]3 years ago
7 0
It would be reasonable to assume that the specific heats of NaOH and HCl solutions are the same as water given that the concentration of these solutions are low about 1 M or less.  Having low concentrations would mean that there is only small amount of particles of HCl or NaOH in the solution so most of the properties of the solution is the same as that of a pure water since less particles can interfere with any process. Specific heat is the amount of heat energy needed per mass in order to be able to raise the temperature by a degree. So, when these particles are present in small amount, it is only the water that would determine the amount of heat needed.
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The boiling point of water is 1000C.
sergejj [24]

Answer:

a) The heat which we supply to water during boiling is used to overcome these forces of attraction between the particles so that they become totally free and change into a gas. This latent heat does not increase the kinetic energy of water particles and hence no rise in temperature takes place during the boiling of water.

b) Steam produces more severe burns than boiling water even though both are at 100oC because steam contains more heat, in the form of latent heat, than boiling water.

Explanation:

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7 0
3 years ago
What is a chemical bond?
nadezda [96]
The answer is A. the attraction between atoms that enables the formation of chemical compounds.
8 0
3 years ago
GIVING BRAINLIEST One mole of hydrogen gas (H2), reacts with one mole of bromine Br2(g) to produce 2 moles of hydrogen bromide g
JulsSmile [24]

Answer:

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

Given, that 1 mole of H_2 gas and 1 mole of Br_2 liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.

H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ

Divide the equation by 2.

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

The equation to show the the correct form to show the standard molar enthalpy of formation:

\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ

4 0
3 years ago
A manganese electrode was oxidized electrically. If the mass of the electrode decreased by 225 mg during the passage of 1580 cou
qwelly [4]

Answer:

<u>Oxidation state of Mn = +4</u>

Explanation:

Atomic mass of Mn = 55g/mol

From Faraday's law of electrolysis,

Electrochemical equivalent = \frac{mass}{charge}

i.e Z = \frac{m}{Q} = \frac{0.225g}{1580C} = 0.0001424 g/C

But Equivalent weight, E = atomic mass ÷ valency  = Z × 96,485

⇒ \frac{55}{valency} = 0.0001424 × 96,485

<u>∴ Valency of Mn = +4</u>

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Answer:

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