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Vikki [24]
3 years ago
13

a normal hemoglobin concentration in the blood is 15g/100ml of blood how many kilogram of hemoglobin are in a person who has 5.5

l of blood
Chemistry
2 answers:
babunello [35]3 years ago
8 0

Answer:

there are 0.825 kg in 5.5 l of blood

Explanation:

First, we have to convert L into mL so we work with the same units

(5.5L)(\frac{1000mL}{1L}) =5500 mL

Using a rule of three we have:

15 g of hemoglobin ------> 100 mL of blood

               x               ------> 5500 mL of blood

x=\frac{(15 g)(5500mL)}{100mL} =825g

As the answer must be in kilogram we must convert grams into kilograms:

(825g)(\frac{1kg}{1000g} )=0.825kg

jok3333 [9.3K]3 years ago
4 0
Using stoichiometry:

5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg
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intrusive igneous rock forms when the cooling of magma takes place fast on the top of the Earth's surface. A. True B. False
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Iron filling's were mixed together with salt crystal's What unique property of iron would be BEST to separate the fillings from
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Answer:

Magnetic property

Explanation:

Iron filling is a magnetic compound, unlike the salt crystals. This means they are attracted by magnets.

To separate a mixture of iron filling s and salt crystals, a magnet should be used to remove the iron fillings from the mixture.

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A steel container with a movable piston contains 2.00 g of helium which was held at a constant temperature of 25 °C. Additional
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Answer: D) 1.00 g

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According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

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where,

V_1 = initial volume of gas  = 2.00 L

V_2 = final volume of gas = 3.00 L

n_1 = initial moles of gas  =\frac{\text {Given mass of helium}}{\text {molar mass of helium}}=\frac{2.00g}{4g/mol}=0.500mol

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

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4 0
3 years ago
A sample of gas occupies 10.0 L at 240°C under a pressure of
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Answer: 1090°C

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where P1 = initial pressure of gas = 80.0 kPa

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P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

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T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

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