All categories

3 years ago

The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?

Chemistry

1 answer:

timurjin [86]3 years ago

5 0

Answer:

150.1 mL

Explanation:

Step 1: Given data

Density of benzene (ρ): 0.879 g/mL

Mass of the sample of benzene (m): 131.9 g

Volume of the sample of benzene (V): ?

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

You might be interested in

Which of these bonds should be classified as ionic?

jolli1 [7]

Answer:

Na-O

Be-S

Explanation:

ionic bonds= metal and non metal

7 0

3 years ago

Read 2 more answers

Cis-platin is a chlorine-containing chemotherapy agent with the formula pt(nh3)2cl2. What is the mass of one mole of cis-platin?

marishachu [46]

Cis-platin is a chemotherapy agent used to treat and kill cancerous cells in patients. One mole of cis-platin has a mass of 300.06 grams/mol. Thus, option B is correct.

<h3>What is a chemotherapy agent?</h3>

A chemotherapy agent is an alkylating agent that is used to treat cancer and is given to reduce the infection or to relieve the symptoms. Cis-platin (Pt(NH₃)₂Cl₂) is one of the chemotherapy agents that treat lung, ovarian, and neck cancer, etc.

The mass of one mole of Cis-platin is calculated as,

Moles = mass ÷ molar mass

Where,

Moles = 1 mole

The molar mass of (Pt(NH₃)₂Cl₂) is calculated as,

195.06 g/mole + 2(17g/mole) + (35.5)2 = 300.06 grams

Substituting values to calculate mass:

Mass = Molar mass × moles

= 300. 06 × 1

= 300.06 grams/mol

Therefore, option B. 300.06 gm/mol is the mass of Cis-platin.

Learn more about chemotherapy agents here:

brainly.com/question/14260402

#SPJ4

Your question is incomplete, but most probably your full question was, Cis-platin is a chlorine-containing chemotherapy agent with the formula pt(nh3)2cl2. What is the mass of one mole of cis-platin?

488.91 g/mol
300.06 g/mol
492.37 g/mol
283.02 g/mol
2860.5 g/mol

8 0

2 years ago

8. Carbohydrates on cell membranes help cells

Natalka [10]

Answer:

Explanation:

5 0

3 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = 0,216 kJ/Kmol

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

5 0

3 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

4 years ago