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3 years ago

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A canister of gases contains oxygen, nitrogen, and hydrogen. Oxygen's partial pressure is 100 torr, nitrogen's partial pressure

is 200 torr, and hydrogen's partial pressure is 50 torr. What is the total pressure in the canister?

1 answer:

olya-2409 [2.1K]3 years ago

7 0

Total pressure = partial pressure for oxygen + partial pressure for nitrogen + partial pressure for hydrogen

100 torr+ 200 torr +50 torr = 350 torr

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An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

BartSMP [9]

Answer:

Odds to be given for an event that either Romance or Downhill wins is 11:4

Explanation:

Given an odd, r = a : b. The probability of the odd, r can be determined by;

Pr(r) = $\frac{a}{b}$ ÷ (

So that;

Odd that Romance will win = 2:3

Pr(R) = $\frac{2}{3}$ ÷ (

= $\frac{2}{3}$ ÷ $\frac{5}{3}$

= $\frac{2}{5}$

Odd that Downhill will win = 1:2

Pr(D) = $\frac{1}{2}$ ÷ (

= $\frac{1}{2}$ ÷ $\frac{3}{2}$

= $\frac{1}{3}$

The probability that either Romance or Downhill will win is;

Pr(R) + Pr(D) = $\frac{2}{5}$ + $\frac{1}{3}$

= $\frac{11}{15}$

The probability that neither Romance nor Downhill will win is;

Pr(neither R nor D) = (1 - $\frac{11}{15}$ )

= $\frac{4}{15}$

The odds to be given for an event that either Romance or Downhill wins can be determined by;

= Pr(Pr(R) + Pr(D)) ÷ Pr(neither R nor D)

= $\frac{11}{15}$ ÷ $\frac{4}{15}$

= $\frac{11}{4}$

Therefore, odds to be given for an event that either Romance or Downhill wins is 11:4

8 0

3 years ago

Consider the halogenation of ethene, where x is a generic halogen: h2c=ch2(g)+x2(g)→h2xc−ch2x(g) you may want to reference (page

KengaRu [80]

Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ

2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ

2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ

2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ

Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction

8 0

3 years ago

Guys plz help me plz

Irina18 [472]

Answer:

The answer is C

Explanation:

3 0

3 years ago

Read 2 more answers

What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl

Kitty [74]

Answer:

$\large \boxed{\text{0.012 mol}}$

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL: 70.

c/mol·L⁻¹: 0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

$\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}$

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

$\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }$

3 0

3 years ago