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2 years ago

Calculate the molar mass of the following:

Chemistry

1 answer:

Westkost [7]2 years ago

7 0

Answer:

a)C2HBrClF3 = 197.35 g/mol

b)C12H14N2CL2 = 229.06g/mol

c)C8H10N4O2 = 194.22g/mol

d) CO(NH2)2=60.07 g/mol

e)C17H35CO2Na = 306.52 g/mol

Explanation:

Molar mass of a compound is equal to the sum of the atomic masses of the constituent elements.

a) C2HBrClF3

$Molar\ mass = 2(At. mass C)+1(at.mass H) +1(At. mass Br) + 1(At.mass Cl) + 3(At.mass F)\\=2(12.01 g/mol) + 1(1.01g/mol)+1(79.90 g/mol) +1(35.45g/mol)+3(18.99g/mol)=197.35g/mol$

b) C12H14N2CL2

$Molar\ mass = 12(C) + 14(H) + 2(N) + 2(Cl)\\\\=12(12.01) + 14(1.01) + 2(14.01) + 2(35.45) = 229.06g/mol$

c) C8H10N4O2

$Molar\ mass = 8(C) + 10(H) + 4(N) + 2(O)\\\\=8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) =194.22g/mol$

d) CO(NH2)2

$Molar\ mass = 1(C) + 1(O) + 2(N) + 4(H)\\\\=1(12.01) + 1(16.00) + 2(14.01)+4(1.01) =60.07 g/mol$

e) C17H35CO2Na

$Molar Mass = 18(C) + H(35) +2(O) + 1(Na)\\\\=18(12.01) + 35(1.01) + 2(16.00) + 1(22.99) =306.52 g/mol$

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The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.

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a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = $\frac{mass of element}{mass of compound}$ (1)

The molecular formula of cisplatin is $Pt(NH_3)_2Cl_2$ .

The atomic weight of the elements in cisplatin is:

Platinum, $Pt = 195.084 u$

Nitrogen, $N = 14.0067 u$

Hydrogen, $H = 1 u$

Chlorine, $Cl = 35.453 u$

The molar mass of $Pt(NH_3)_2Cl_2$ = $195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453)$ = $300.00 g/mol$

The mass of each element calculated using formula (1):

- Platinum, $Pt$ %

$\frac{195.084}{300.00} \times 100 = 65.23$ %.

- Nitrogen, $N$ %

$\frac{2\times 14.0067}{300.00} \times 100 = 9.34$ %

- Hydrogen, $H$ %

$\frac{6\times 1}{300.00} \times 100 = 2.0$ %

- Chlorine, $Cl$ %

$\frac{2\times 35.453}{300.00} \times 100 = 23.63$ %

b. The given reaction of cisplatin is:

$K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)$

According to the balanced reaction, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ .

Now, calculating the number of moles of $K_2PtCl_4$ in 100.0 g.

Number of moles = $\frac{given mass}{Molar mass}$

Molar mass of $K_2PtCl_4$ = $2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol$

Number of moles of $K_2PtCl_4$ = $\frac{100 g}{415.093 g/mol} = 0.241 mole$ .

Since, 1 mole of $K_2PtCl_4$ gives 1 mole of $Pt(NH_3)_2Cl_2$ . Therefore, mass of cisplatin is:

$0.241 mole\times 300 g/mol = 72.3 g$

For mass of $KCl$ :

Molar mass of $KCl$ = $39.0983 + 35.453 = 74.55 g/mol$

Since, 1 mole of $K_2PtCl_4$ gives 2 mole of $KCl$ . Therefore, mass of $KCl$ is:

$0.241 mole\times 74.55 g/mol\times 2 = 35.93 g$

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3 years ago

What is the [H+] of a solution with a pH of 9.40?

LiRa [457]

Answer:

a) 3.98 x 10^-10

Explanation:

Hello,

In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:

$pH=-log([H^+])$

Hence, solving for the concentration of hydronium:

$[H^+]=10^{-pH}=10^{-9.40}\\$

$[H^+]=3.98x10^{-10}M$

Therefore, answer is a) 3.98 x 10^-10

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