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2 years ago

Helpppppppppppppppppppppppppppp

Chemistry

2 answers:

worty [1.4K]2 years ago

6 0

Answer:

evaporation or water runoff that's my tips

rain or thunder storm

Dvinal [7]2 years ago

3 0

Answer:

<h2>condensation </h2>

Explanation:

Condensation is a process in which the evaporated liquid substance such as water vapor gets turned back to its liquid form water.

So, condensation  is the process in which water vapor changes to liquid .

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What are two variables that affect kinetic energy

zubka84 [21]

The answer for the following questions is explained below.

Explanation:

The two variables that affect kinetic energy are:

mass and
velocity

velocity - The faster an object moves,the more the kinetic energy it has.

mass - Kinetic energy increases as mass increases

The kinetic energy of an object depends on both its mass and its velocity

Kinetic energy increases as mass increases

For example,think about rolling a bowling ball and a golf ball down a bowling lane at same velocity

Here,the bowling ball has more mass than the golf ball

Therefore you use more energy to roll the bowling ball than to roll the golf ball

The bowling ball is more likely to knock down the pins because it has more kinetic energy than the golf ball

7 0

2 years ago

1)petroleum and natural gas are formed in the__________

ale4655 [162]

1. In the remains of organisms such as plants and algae.
2. Air, peace of mind know knowleged and water
3. The Carboniferous
4. Is a colorless , odorless gass that is very flammable.

6 0

2 years ago

Please help and show work, i will mark you brainlest

Natali [406]

Answer:

7x-2(3-4)-2

Pls help

Explanation:

8 0

2 years ago

How much heat is lost when 575 grams molten iron at 1825 k becomes solid iron at 293 k? The melting point or iron is 1811 k.

Galina-37 [17]

Answer:

146 kJ

Explanation:

There are two heat flows in this question.

Heat lost on cooling + heat lost on solidifying = 0

q₁ + q₂ = 0

mCΔT + nΔHsol = 0

Data:

m = 575 g

C = 0.449 J·K⁻¹g⁻¹

T_i = 1825 K

T_f = 1811 K

ΔHsol = -13.8 kJ·mol⁻¹

Calculations:

(a) Heat lost on cooling

ΔT = T_f - T_i = 1811 K - 1825 K = -14 K

q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ

(b) Heat lost on solidifying

$n = \text{575 g} \times \dfrac{\text{1 mol}}{\text{55.84 g}} = \text{10.30 mol}\\\\q_{2} = n\Delta_{\text{sol}}H = \text{10.30 mol} \times \dfrac{\text{-13.8 kJ}}{\text{1 mol}}= \text{-142.1 kJ}$