HELP NOW BRAINLIST AND 15 POINTS!!!!!! PLEASE HELPPP

Chemistry

2 answers:

Varvara68 [4.7K]3 years ago

7 0

Answer:

1. Chemical change

2. It is creating a new substance

sorry for typeos or if wrong and for not answering all

ycow [4]3 years ago

3 0

Answer:

sis why u doing school at night Lol.Save it for me tomorrow

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A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.

Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L$

Therefore, the new volume of the air is, 6.83 L

6 0

3 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$