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2 years ago

Blood has a mass-volume percent of NaCl of 0.9%. What mass (g) of NaCl is present in a liter of blood

Chemistry

1 answer:

quester [9]2 years ago

6 0

The mass (g) of NaCl that is present in a liter of blood is 0.347g.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the decimal mass-volume of the substance by its molar mass.

According to this question, Blood has a mass-volume percent of NaCl of 0.9%. This means that 0.9/100 = 0.009g of NaCl is present in each 100g of blood.

Molar mass of NaCl = 23 + 35.5 = 38.5g/mol

Mass of NaCl in a liter of blood = 38.5 × 0.009

Mass of NaCl in a liter of blood = 0.347g of NaCl.

Therefore, The mass (g) of NaCl that is present in a liter of blood is 0.347g.

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How does insulin bind to cells and the mechanism involved in triggering the cells to take in glucose

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Explanation:

Once blood glucose levels increase, pancreatic insulin migrates into a fat cell via the blood stream. Insulin then binds in the plasma membrane of the cell to an Insulin Receptor (IR). Through autophosphorylation, phosphate groups are then added to the IR, causing GLUT4 molecules to come to the cell's surface.

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3 years ago

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

melamori03 [73]

Answer: The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For nickel:

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

For nickel (II) oxide:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

For aluminium:

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0

3 years ago

90.00 ml of 0.25 M calcium hydroxide are required to titrate 100.00 ml of hydrochloric acid. What is the molarity of the HCI?

bezimeni [28]

Answer:

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Explanation:

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3 years ago

A 10.0-g sample of sodium chloride was placed in 10.0 g of water. If 3.85 g of Cl2 was obtained, what was the percent yield of C

suter [353]

Answer:

63.53% yield

Explanation:

The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2

First we must find the limiting reactant

From NaCl we can only produce 6.06 grams of Cl2 in theory

From H20 we can only produce 38.995 grams in theory

so we know NaCl is the limiting

% yield is (Actual/Theoretical) x100 so

(3.85/6.06)x100= 63.53% yield

7 0

2 years ago

if 15 mol C is mixed with 10 mol O2 which reactant is the limiting reactant? Which reactant would be the excess reactant?

ioda

Answer:

C is the excess reactant.

Explanation:

Reaction is C + O2 --> CO2

1mol of C required to react with 1mol O2

Therefore 15 - 10 = 5moles of C will be in excess

6 0

3 years ago