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bearhunter [10]
3 years ago
14

Which of tge following is an example of a chemical change

Chemistry
1 answer:
Effectus [21]3 years ago
8 0
Α.fℓσυя αи∂ ѕυgαя αяє ѕιfтє∂ тσgєтнєя ιитσ α мєαѕυяιиg ¢υρ
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Imagine that you and your friend are at the beach. You have your chairs and umbrella set up in the perfect spot near the water s
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High tides and low tides are caused by the moon. The moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the moon and the side farthest from the moon. ... When you're not in one of the bulges, you experience a low tide.

8 0
3 years ago
6. A sample of a gas at 77°C and 1.33 atm occupies a volume of 50.3 L. How many moles of the gas are present? (Hint: Since moles
Nina [5.8K]

The number of moles of the gas at 77°C and 1.33 atm occupies a volume of 50.3 L is 2.35 moles. It can found with the help of Ideal gas equation.

<h3>What is Ideal Gas equation ? </h3>

The ideal gas equation is formulated as : PV = nRT.

In this equation, P refers to the pressure of the ideal gas, V is the volume of the ideal gas, n is the total amount of ideal gas that is measured in terms of moles, R is the universal gas constant, and T is the temperature.

Given ;

  • Pressure = 1.33 atm
  • Volume = 50.3 ltr
  • Temperature = 77 (+273 k) = 350K

We know ;

  • Gas constant (R) = 0.081 L atm/mol K

Formula used ;

n = PV / RT

n = 1.33 x 50.3 / 0.081 x 350k

  =  2.35 moles.

Hence, The number of moles of the gas at 77°C and 1.33 atm occupies a volume of 50.3 L is 2.35 moles

Learn more about Ideal Gas here ;

brainly.com/question/27922399

#SPJ1

3 0
3 years ago
40 POINTS!!! Ice at 0.0 degrees C is combined with 50.0g of water at 75.0 degrees C. Calculate the grams of ice present initiall
nydimaria [60]

Answer:

mi = 31.28 g

Explanation:

According to the law of conservation of energy:

Heat\ Gain\ by\ Ice = Heat\ Lost\ by\ Water\\m_iL = m_wC\Delta T\\

where,

mi = mass of ice = ?

L = Latent heat of fusion of ice = 334 J/g

mw =  mass of water = 50 g

C = specific heat of water = 4.18 J/g.°C

ΔT = change in temperature of water = 75°C - 25°C = 50°C

Therefore,

m_i(334\J/g) = (50\ g)(4.18\ J/g.^oC)(50^oC)

<u>mi = 31.28 g</u>

8 0
3 years ago
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