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UNO [17]
3 years ago
6

Select the correct equalities for the units shown.

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer

1. 1cm=10mm, 1kg=1000g and 12in=1 ft

2. 50mm=5cm

    48 in =4ft

     36ft=12 yd

     2.5m=250cm

     1000mL= 1 L

Explanation

if 1cm=10mm then 5cm=?

5×10=50mm

if 12 in =1ft then 48in=?

48/12 =4 ft

if 1m=100cm, then 2.5m=?

2.5×100= 250cm

and when know from capacity that 1000mL is equivalent to 1 liter.

Allushta [10]3 years ago
7 0

Answer:

1. 50mm, 2. 4ft, 3. 4ft, 4. 250cm, 5. 1L

Explanation:

1. 1cm = 10mm

5cm = y mm = y = (5*10)/1 = 50mm

2. 12in = 1ft

48in = yft

y ft = (48*1) / 12 = 4ft

3. 1ft = 3yd

Z ft = 12yd

Z ft = (12*1)/3 = 4ft

4. 1m = 100cm

2.5m = x cm

X = (100*2.5) / 1 = 250cm

5. 1mL = 1*10⁻³L

1000mL = z L

z L = (1000 * 1*10⁻³) / 1

Z L = 1L

1000mL = 1L.

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Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressure
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The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of O_2 and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = n_1=\frac{51.8}{32 g/mol}=1.619 mol

Moles of helium gas = n_2=\frac{33.1 g}{4 g/mol}=8.275 mol

Total moles of gas = n_1+n_2=(1.619 +8.275 ) mole=9.894 mol

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

P=\frac{nRT}{V}

=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}

p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm

Partial pressure of the helium gas

p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}

p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm

6 0
3 years ago
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