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yarga [219]
3 years ago
14

Fill in the blanks with the correct types of energy.

Chemistry
2 answers:
liq [111]3 years ago
4 0

Answer:

I would say C but I am just in 7th grade soo

rodikova [14]3 years ago
4 0
Yup it’s C :)) oh no i need to add more characters
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Need answer quick please
Nikolay [14]
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using avogadro’s law
6 0
3 years ago
A sample of gas occupies a volume of 67.5 mL . As it expands, it does 131.0 J of work on its surroundings at a constant pressure
baherus [9]

Answer:

The final volume V2=1.3175L

Explanation:

between work ( w), pressure ( P ) and volume ( V ) is the following:

w=−PΔV

where,

ΔV=V2−V1

It was stated that the gas is expanding, then the work is done by the system and it is of a negative value .

Note that work, should be expressed in 1L⋅atm=101.3J

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

4 0
4 years ago
Bicycles that have been left outside in the summer will rust faster than bicycles kept inside. What causes the bicycles that are
professor190 [17]

Bikes that are kept outside are not used as often, speeding up the chemical reaction of metal rusting.
Heat energy from the sun slows down the chemical reaction of the metal rusting.
Moisture and oxygen cause oxidation, which speeds up the chemical reaction of the metal rusting.
Wind energy outside speeds up the chemical reaction of the metal rusting.
8 0
3 years ago
A small bubble rises from the bottom of a lake where the temperature and pressure are 4.0 C and 3.0 atm, to the water’s surface,
Varvara68 [4.7K]

Answer:

The final volume of the bubble is 7.13 mL.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 3 atm

P_2 = final pressure of gas = 0.95 atm

V_1 = initial volume of gas = 2.1 mL=0.0021 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 4^oC=273.15+4K=277.15 K

T_2 = final temperature of gas = 25^oC=273.15+25 k=298 .15 K

Now put all the given values in the above equation, we get:

\frac{3 atm\times 0.0021 L}{277.15 K}=\frac{0.95 atm\times V_2}{298.15 K}

V_2=\frac{3 atm\times 0.0021 L\times 298.15 K}{277.15 K\times 0.95 atm}=0.00713 L = 7.13 mL

The final volume of the bubble is 7.13 mL.

5 0
3 years ago
a 20.0mL sample of 0.15M hydrochloric acid (HCI) is needed to neutralize a 10.0mL sample of potassium hydroxide (KOH). what is t
Masteriza [31]

Answer: .75 M

Explanation:

8 0
2 years ago
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