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2 years ago

HELP ME OUT MY LOVELY CHEMISTRY CHADS I WILL GIVE YOU A KISS

Chemistry

1 answer:

vitfil [10]2 years ago

6 0

Answer:

300K.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 300L

Initial temperature (T1) = 200K

Final volume (V2) = 450L

Final temperature (T2) =..?

Since the pressure is constant, the gas is obeying Charles' law.

Using the Charles' law equation, we can obtain the new temperature of the gas as follow:

V1/T1 = V2/T2

300/200 = 450/T2

Cross multiply to express in linear form

300 x T2 = 200 x 450

Divide both side by 300

T2 = (200 x 450)/ 300

T2 = 300K

Therefore, the new temperature of the gas is 300K.

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Exactly one mole of an ideal gas is contained in a 2.00-liter container at 1,000 K. What is the pressure exerted by this gas?

levacccp [35]

Use PV =nRT

so P = nRT/V

= 1 mole(0.08205 L atm/K mol)(1000K) / 2 L

= 41 atm

7 0

2 years ago

Read 2 more answers

A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is th

lara [203]

Answer is: the partial pressure of the helium gas is 0.158 atm.

p(mixture) = 0.48 atm; total pressure.

m(H₂) = 1.0 g; mass of hydrogen gas.

n(H₂) = m(H₂) ÷ M(H₂).

n(H₂) = 1.0 g ÷ 2 g/mol.

n(H₂) = 0.5 mol; amount of hydrogen.

m(He) = 1.0 g; mass of helium.

n(He) = 1 g ÷ 4 g/mol.

n(He) = 0.25 mol; amount of helium.

χ(H₂) = 0.5 mol ÷ 0.75 mol.

χ(H₂) = 0.67; mole fraction of hydrogen.

χ(He) = 0.25 mol ÷ 0.75 mol.

χ(He) = 0.33; mole fraction of helium.

p(He) = 0.33 · 0.48 atm.

p(He) = 0.158 atm; the partial pressure of the helium gas.

8 0

3 years ago

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is

Charra [1.4K]

Answer:

$[H^{+}] = 0.761 \frac{mol}{L}$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

$pH = 0.119$

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

$n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)$

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

$n_{H^{+} } from HCl = (5.00)(0.093)$

$n_{H^{+} } from HCl = 0.465 mol$

$n_{H^{+} } from HNO_{3} = (8.00)(0.037)$

$n_{H^{+} } from HNO_{3} = 0.296 mol$

$n_{H^{+}(total) } = 0.296 + 0.465$

$n_{H^{+}(total) } = 0.761 mol$

For molar concentration of hydrogen ions:

$[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}$

$[H^{+}] = \frac{0.761}{1.00}$

$[H^{+}] = 0.761 \frac{mol}{L}$

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

$K_{w} = [H^{+} ][OH^{-} ]$

$[OH^{-}]=\frac{Kw}{[H^{+}] }$

$[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }$

$[OH^{-}]=1.33X10^{-14}\frac{mol}{L}$

The pH of the solution can be measured by the following formula:

$pH = -log[H^{+} ]$

$pH = -log(0.761)$

$pH = 0.119$

5 0

2 years ago

Able 1 Cell Type Operating Cell Potential for Commercial Batteries, E (V) Lithium-iodine Zinc-mercury +2.80 +1.35 Table 2 Standa

IrinaK [193]

Answer:

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. ( B )

During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not. ( C )

Explanation:

<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>

During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.

Given the relationship below,

Δ G = -nFE

E = emf of cell , G = free energy.

This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable

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2 years ago

Why is activation energy required to start some reactions?

maxonik [38]

The activation energy is the minimum required energy in order to start a reaction. It is the barrier to break to enable reactants to form products. It is not important only for some reactions but to all reactions. This energy should be supplied or reached first before we observe a reaction to happen.

4 0

3 years ago