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8_murik_8 [283]
3 years ago
5

Please help me!! I’m confused and this teacher doesn’t know how to teach.

Chemistry
1 answer:
Marianna [84]3 years ago
8 0

Answer:

chemical change bcoz it cannot be in the past form

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2 years ago
How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group
OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

  • P= 1.00 atm
  • V= 150 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

mass=\frac{6.69 moles*70.9 grams}{1 mole}

mass= 474.321 grams ≅ 474 grams

<u><em>474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C</em></u>

4 0
3 years ago
Atoms of elements in the same group have the same number of
kherson [118]

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electrons (can i have brainliest)

Explanation:

7 0
3 years ago
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Explain the properties of the substance around us depend on the atoms and molecules that compose them
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7 0
3 years ago
The specific heat of liquid bromine is 0.226 J/g-K. How much heat (J) is required to raise the temperature of 10.0 mL of bromine
Viefleur [7K]

Answer:

16.2 J

Explanation:

Step 1: Given data

  • Specific heat of liquid bromine (c): 0.226 J/g.K
  • Volume of bromine (V): 10.0 mL
  • Initial temperature: 25.00 °C
  • Final temperature: 27.30 °C
  • Density of bromine (ρ): 3.12 g/mL

Step 2: Calculate the mass of bromine

The density is equal to the mass divided by the volume.

ρ = m/V

m = ρ × V

m = 3.12 g/mL × 10.0 mL

m = 31.2 g

Step 3: Calculate the change in the temperature (ΔT)

ΔT = 27.30 °C - 25.00 °C = 2.30 °C

The change in the temperature on the Celsius scale is equal to the change in the temperature on the Kelvin scale. Then, 2.30 °C = 2.30 K.

Step 4: Calculate the heat required (Q) to raise the temperature of the liquid bromine

We will use the following expression.

Q = c × m × ΔT

Q = 0.226 J/g.K × 31.2 g × 2.30 K

Q = 16.2 J

7 0
3 years ago
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