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Rainbow [258]
2 years ago
15

Calculate the molarity of a solution prepared by dissolving 1.495 moles of lioh in enough water to give a final volume of 750. m

l.
Chemistry
1 answer:
zhenek [66]2 years ago
4 0
Molarity can be defined as the number of moles of substance dissolved in 1 L of solution.
In the given question ,
number of LiOH moles - 1.495 mol
Dissolved volume - 750 mL
molarity is calculated for 1 L = 1000 mL
In 750 mL - 1.495 mol of LiOH is dissolved
Therefore in 1000 mL - 1.495 mol / 750 mL x 1000 = 1.99 mol

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7. A gas ______.
enot [183]

Answer:

C

Explanation:

A gas is a state of matter that has no fixed shape and no fixed volume gases have lower density than other states of matter such as solids and liquids.

7 0
3 years ago
What is the energy of a radio wave whose frequency is 6.2 x 106 Hz?
slava [35]

Answer:

657.2 Hz

Explanation:

7 0
2 years ago
Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11
otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = \frac{2400 g}{890 g/mol}=2.6966 mol

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give  ;

\frac{110}{2}\times 2.6966 mol=148.31 mol of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0
2 years ago
The name of the following molecular compound: P203
topjm [15]
B Phosphate trioxide
5 0
2 years ago
At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

3 0
2 years ago
Read 2 more answers
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