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Mkey [24]
2 years ago
15

This download is my science worksheet. It's about different types of energy in a situation, like examples. btw this is 7th grade

science

Chemistry
1 answer:
scoray [572]2 years ago
8 0

All types of energy can be resumed into two basic types of energy which include kinetic energy and potential energy.

<h3>What is kinetic energy?</h3>

Energy is the ability to perform a given work. Kinetic energy is energy in movement, whereas potential energy is stored energy.

For example, plant photosynthesis makes reference to chemical energy (potential energy), popcorn makes reference to thermal energy, etc.

In conclusion, all types of energy can be resumed into two basic types of energy which include kinetic energy and potential energy.

Learn more about kinetic energy here:

brainly.com/question/25959744

#SPJ1

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A dull metal object has a density of 8.8 G/ML and a volume of 20 ML calculate the mass
Alex

Answer:

Mass = 0.000176 gram

Steps:

m =  V × ρ

=  20 milliliter × 8.8 gram/cubic meter

=  2.0E-5 cubic meter × 8.8 gram/cubic meter

=  0.000176 gram

Explanation:

8 0
3 years ago
Consider the half reaction below.
aksik [14]

Answer:

It's B on Edge

Explanation:

8 0
3 years ago
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A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an
laila [671]

The specific heat of metal is c = 3.433 J/g*⁰C.

<h3>Further explanation</h3>

Given

mass of metal = 68.6 g

t metal = 100 °C

mass water = 84 g

t water = 20 °C

final temperature = 52.1  °C

Required

The specific heat

Solution

Heat can be formulated :

Q = m.c.Δt

Q absorbed by water = Q released by metal

84 x 4.184 x (52.1-20)=68.6 x c x (100-52.1)

11281.738=3285.94 x c

c = 3.433 J/g*⁰C.

7 0
3 years ago
A 1.00 g sample of n-hexane (C6H14) undergoes complete combustion with excess O2 in a bomb calorimeter. The temperature of the 1
Nadya [2.5K]

Answer:

-5,921x10⁶J/mol

Explanation:

Internal energy change (ΔU) for the reaction of combustion in the bomb calorimeter is:

ΔU = q calorimeter + q solution

Where:

q calorimeter is Ccal×ΔT (Ccal=4042J/°C) and (ΔT is 29,30°C-22,64°C=<em>6,66°C</em>)

q solution is c×m×ΔT (c= 4.184 J/g°C), (m=1502g H₂O), (ΔT is 29,30°C-22,64°C=<em>6,66°C</em>)

Replacing:

ΔU = 26920J + 41854J = <em>68774 J</em>

This energy is per g of n-hexane, now, per mole of n-hexane:

\frac{68774J}{1gHexane} *\frac{86,1g}{1mol}= -<em>5,921x10⁶J/mol</em>

<em>-negative because the energy is produced-</em>

I hope it helps!

5 0
4 years ago
Read 2 more answers
If sea water has a density of 1.025 kg/L and 3.5% of the mass in sea water is salt, determine the mass of salt dissolved in 500
Alja [10]

68 kg. There are 58 kg salt in 500 gal seawater.

<em>Step 1.</em> Convert gallons to litres

1 US gal = 3.79 L (1 Imp gal = 4.55 L)

<em>Step 2</em>. Find the volume of the seawater

Volume = 500 gal × (3.79 L/1 gal) = 1895 L

<em>Step 3</em>. Find the mass of the seawater

Mass = 1895 L × (1.025 kg/1 L) = 1942 kg

<em>Step 4</em>. Find the mass of the salt

Mass of salt = 1942 kg seawater × (3.5 kg salt/100 kg seawater) = 68 kg salt

6 0
4 years ago
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