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sashaice [31]
3 years ago
12

What is the pressure, in mmHg, of a 4.00 g sample of O2 gas, which has a temperature of 37.0 °C, and a volume of 4400 mL?

Chemistry
1 answer:
nadya68 [22]3 years ago
5 0

Answer:

549.48 mmHg

Explanation:

We'll begin by calculating the number of mole of oxygen in 4 g. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ = 4 g

Mole of O₂ =?

Mole = mass /molar mass

Mole of O₂ = 4/32

Mole of O₂ = 0.125 mole

Next, we shall convert 37.0 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T (°C) = 37.0 °C

T(K) = 37.0 °C + 273

T(K) = 310 K

Next, we shall convert 4400 mL to L.

1000 mL = 1 L

Therefore,

4400 mL = 4400 mL × 1 L / 1000 mL

4400 mL = 4.4 L

Next, we shall determine the pressure. This can be obtained as follow:

Number of mole (n) = 0.125 mole

Temperature (T) = 310 K

Volume (V) = 4.4 L

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

PV = nRT

P × 4.4 = 0.125 × 0.0821 × 310

Divide both side by 4.4

P = (0.125 × 0.0821 × 310) / 4.4

P = 0.723 atm

Finally, we shall convert 0.723 atm to mmHg.

1 atm = 760 mmHg

Therefore,

0.723 atm = 0.723 atm × 760 mmHg / 1 atm

0.723 atm = 549.48 mmHg

Thus, the pressure is 549.48 mmHg

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Answer:

Density = 4.06 * 10^14

Explanation:

Givens

r = 1 * 10^-13 cm

m = 1.7 * 10^-24 grams

Formula

V = 4/3 * pi * r^3

D = m/V

Solution

Put the givens into the formula.

V = (4/3) 3.14 * (1 * 10^-13)^3          Find the value for r^3

V = (4/3) 3.14 * 1 * 10^-39               Find the volume. Indicate divide by 3

V = 1.256 10^-38/3

V = 4.1867 * 10^-39

Density = m / v

Density = 1.7*10^-24/4.1867*10^-39  Do the division

Density = .406 * 10^(-24 + 39)

Density = .406 * 10^15 grams/ cm^3

Answer: Density = 4.06 * 10^14

5 0
2 years ago
HELP ASAP Will Give Brainlist
a_sh-v [17]

The correct answer is actually

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3 0
3 years ago
Calculate the volume in mL of 1 mole of pure ethanol
Tamiku [17]
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4 0
4 years ago
Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days.
8090 [49]

Answer:

11.9g remains after 48.2 days

Explanation:

All isotope decay follows the equation:

ln [A] = -kt + ln [A]₀

<em>Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope</em>

We can find k from half-life as follows:

k = ln 2 / Half-Life

k = ln2 / 27.7 days

k = 0.025 days⁻¹

t = 48.2 days

[A]  = ?

[A]₀ = 39.7mg

ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]

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8 0
3 years ago
Why was 1,2 dichlorobenzene used as the solvent for the diels alder reaction we performed in the lab?
pav-90 [236]

Answer:

1,2 dichlorobenzene was used as the solvent for the diels alder reaction: <em>because the co elimination part of the reaction needs high temp and a high boiling point solvent such as 1,2 dichlorobenzene</em>

Explanation:

Diels-Alder Reaction is a useful synthetic tool to prepare cyclohexane rings. It is a process, which occurs in a single step that consists of a cyclic redistribution of its electrons. The two reagents are bond together through a cyclic transition state in which the two new C-C bonds are formed at the same time. For this to occur, most of the time, it is necessary a high temperature and high-pressure conditions. Since 1,2 dichlorobenzene has a boiling point of 180ºC is a good solvent for this type of reactions.

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