All categories

3 years ago

How many moles are present in 10.5 grams of NaCl?

Chemistry

2 answers:

konstantin123 [22]3 years ago

6 0

Answer : The number of moles present in 10.5 grams of NaCl are, 0.179 moles.

Solution : Given,

Mass of NaCl = 10.5 g

Molar mass NaCl = 58.5 g/mole

Formula used :

$\text{Moles of NaCl}=\frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}}$

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

$\text{Moles of NaCl}=\frac{10.5g}{58.5g/mole}=0.179moles$

Therefore, the number of moles present in 10.5 grams of NaCl are, 0.179 moles.

snow_lady [41]3 years ago

4 0

Mr of NaCl = 58.5

Mole = mass÷ Mr..

Mole od NaCl = 10.5 ÷ 58.5 = 0.179

You might be interested in

The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

ai) Rate law, $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

aii) Rate constant, k = 1.25

b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

$CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)$

If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

k is constant for all the stages

Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction = a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

3 0

3 years ago

List the following elements in order of increasing number of valence electrons: C, CI, As, Na, He.

lorasvet [3.4K]

B is the correct answer

4 0

3 years ago

What is the predominant intermolecular force in CBr4

vovikov84 [41]

CBr4 is a symmetric tetrahedral molecule so it will be non-polar.

3 0

3 years ago

A 1.42-g sample of a pure compound with formula m2so4 was dissolved in water and treated with an ex- cess of aqueous calcium chl

solong [7]

The reaction between $m_{2}SO_{4}$ with calcium chloride can be shown as- $m_{2}SO_{4}$ + $CaCl_{2}$ → $CaSO_{4}$ ↓+2mCl. The molecular weight of $CaSO_{4}$ is 136.14g. The weight of sulfate ion is 96.06g. The molecular weight of $m_{2}SO_{4}$ = (2×m + 96.06). From the reaction we can see that 1 mole of calcium chloride reacts with 1 moles $m_{2}SO_{4}$ to produce 1 mole of calcium sulfate. Now 1.36g of calcium sulfate is equivalent to 1.36/136.14=9.989× $10^{-3}$ moles of calcium sulfate.