3 years ago

Plz help .....................

Chemistry

1 answer:

Pavel [41]3 years ago

6 0

Answer:

1. conduction

2. radiation

3.convection

4. conduction?

5. radiation

6. convection

7. convection

8. convection

9. radiation

10.radiation

(sorry if they aren't all correct)

You might be interested in

Calculate the change in temperature that occurs when 3.50 x 10^-2kg of copper heated by 635 J of energy. Ccu = 0.385 j/gxC

vovikov84 [41]

Answer:

The change in T° is 47.1 °C

Explanation:

Calorimetry formula to solve this:

Q = m . C . ΔT

We replace the data gien:

635 J = 3.50×10⁻² kg . 0.385 J/g°C . ΔT

In units of C, we have g and the mass (m) is in kg. So let's convert it from kg to g → 3.50×10⁻² kg . 1000 g / 1kg = 35 g. Now we can determine the ΔT:

635 J = 35 g . 0.385 J/g°C . ΔT

635 J / 35 g . 0.385 J/g°C = ΔT

47.1°C = ΔT

6 0

3 years ago

Read 2 more answers

Salt water can be desalinated by distillation. How much energy is needed to convert 175 g of salt water at 28.0 °C to water vapo

Vinil7 [7]

Answer:

$\Delta H=4.63x10^{5}J=463kJ$

Explanation:

Hello,

In this case, we find the following states:

a. Liquid salt water at 28.0 °C.

b. Liquid salt water at 102.5 °C.

c. Vapor salt water at 102.5 °C.

The first process (1) is to heat the liquid water from 28.0 °C to 102.5 °C and the second one (2) to vaporize the liquid salt water. In such a way, each process has an amount of energy that when added, yields the total energy for the process as shown below:

$\Delta H=\Delta H_1+\Delta H_2\\\Delta H=mCp\Delta T+m\Delta _VH\\\Delta H=175g*5.19\frac{J}{g*K}*(375.65K-301.15K)+175g*2.26\frac{kJ}{g} *\frac{1000J}{1kJ} \\\Delta H=4.63x10^{5}J=463kJ$