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Mila [183]
3 years ago
14

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]3 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.

Learn more at: brainly.com/question/15743584?referrer=searchResults

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How many moles of sodium chloride can react with 18.3 liters of fluorine gas at 1.2 atmospheres and 299 Kelvin?
My name is Ann [436]

Answer:

1.79 mol.

Explanation:

  • For the balanced reaction:

<em>2NaCl + F₂ → 2NaF + Cl₂. </em>

It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.

  • Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 1.2 atm).

V is the volume of the gas in L (V = 18.3 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (299 K).

∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.

  • Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:

<em><u>Using cross multiplication:</u></em>

2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.

??? mol of NaCl is needed to react with → 0.895 mol of F₂.

∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.

4 0
3 years ago
Consider the reaction CaCN2 + 3 H2O → CaCO3 + 2 NH3 . This reaction has a 75.6% yield. How many moles of CaCN2 are needed to obt
kherson [118]

Answer: Thus 0.724 mol of CaCN_2 are needed to obtain 18.6 g of NH_3

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} NH_3=\frac{18.6g}{17g/mol}=1.09moles

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

According to stoichiometry :

2 moles of NH_3 are produced by = 1 mole of CaCN_2

Thus 1.09 moles of NH_3 will be produced by =\frac{1}{2}\times 1.09=0.545moles  of CaCN_2

But as yield of reaction is 75.6 %, the amount of CaCN_2 needed is =\frac{0.545}{75.6}\times 100=0.724

Thus 0.724 mol of CaCN_2 are needed to obtain 18.6 g of NH_3

3 0
3 years ago
Water flows from the bottom of a large tank where the pressure is 100 psig through a pipe to a turbine which produces 5.82 hp. T
Marianna [84]

Explanation:

Bernoulli equation for the flow between bottom of the tank and pipe exit point is as follows.

   \frac{p_{1}}{\gamma} + \frac{V^{2}_{1}}{2g} + z_{1} = \frac{p_{2}}{\gamma} + \frac{V^{2}_{2}}{2g} + z_{2} + h_{f} + h_{t}

    \frac{(100 \times 144)}{62.43} + 0 + h[tex] = [tex]\frac{(50 \times 144)}{(62.43)} + \frac{(70)^{2}}{2(32.2)} + 0 + 40 + 60

                          h = \frac{(50 \times 144)}{(62.43)} + \frac{(70)^{2}}{2(32.2)} + 40 + 60 - \frac{(100 \times 144)}{(62.43)}

                            = 60.76 ft

Hence, formula to calculate theoretical power produced by the turbine is as follows.

                                 P = mgh

                                     = 100 \times 60.76

                                     = 6076 lb.ft/s

                                     = 11.047 hp

Efficiency of the turbine will be as follows.

                \eta_{t} = \frac{P_{actual}}{P_{theoretical}} × 100%

                                = \frac{5.82}{11.047} \times 100%                      

                                = 52.684%

Thus, we can conclude that the efficiency of the turbine is 52.684%.

4 0
3 years ago
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