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Nataly_w [17]
3 years ago
10

Why does salt dissollve in water act as an antifreeze

Chemistry
2 answers:
Svetlanka [38]3 years ago
7 0
It dissolves the hydrogen in water
AnnyKZ [126]3 years ago
3 0
Salt disrupt the bonds of hydrogen in water. This makes the freezing point of water decrease. In other words, as the saltiness increases, the freezing point decreases. This is why salt dissolved in water acts as antifreeze. Best of luck to you, my buddy.
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What is the temperature change on a 75.0 gram sample of mercury if 480.0 cal of heat are added to it? The specific heat of mercu
Vesna [10]

Answer:

\Delta T=194^oC

Explanation:

Hello,

In this case, the relationship between the mass, heat, temperature and heat capacity id given by:

Q=mCp\Delta T

In such a way, the temperature change results:

\Delta T=\frac{Q}{mCp}=\frac{480.0cal}{75.0g*0.033cal/(g^oC)}  \\\\\Delta T=194^oC

Clearly it is a positive change which means the temperature increases as heat is added.

Regards.

7 0
3 years ago
5. Using the terms: metal and nonmetal, describe what types of atoms make up each type of<br> bond.
Elis [28]

Answer:

See the answer below, please.

Explanation:

The bonds formed between metals and nonmetals are called ionics. These occur between atoms with electronegativity difference. Example: NaCl (Sodium Chloride)

Instead, covalent bonds are formed between two nonmetals (one or more electron pairs are shared). Example: H202 (hydrogen peroxide).

In the case of metal formed bonds, they are called metallic.

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3 years ago
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3 years ago
A sample of hydrogen gas is placed in a 0.500 L container at 295K. The gas pressure is 1.442 bar. How many moles of H2 gas are i
steposvetlana [31]

Answer:

0.0294

Explanation:

5 0
3 years ago
Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0
3 years ago
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