Alice has type A blood and her husband Mark has Type B blood

Why is the collision theory considered to be well-established and highly reliable?

hodyreva [135]

BECAUSE IT NEVER BREAKES

5 0

3 years ago

Why can't the temperature of an object alone

lukranit [14]

Answer:

It cant

Explanation:

It can not tell because the size of a human and an object to 2 different sizes. Therefor since they are 2 different sizes they can handle different amounts of temperature.

4 0

2 years ago

Write equations for the initiation and propagation steps for the formation of dichloromethane by free radical

olya-2409 [2.1K]

Name of type of mechanism initiation step first propagation step second propagation step(ii) write an overall equation for the formation of dichloromethane from ... Best Answer: i) This is a free-radical substitution mechanism.

7 0

2 years ago

When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Yanka [14]

Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂ → CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

C : CO₂

1 : 1

1.2 : 1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2 mol × 44 g/mol

Mass = 52.8 g

8 0

2 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.

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

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.

Learn more: brainly.com/question/4400908

7 0

2 years ago