A gas mixture contains 1.61 moles of hydrogen and 2.31 moles of oxygen. What is the mole fraction of oxygen?

Radioactive manganese-52 decays with a half-life of 5.6 days. A chemist obtains a fresh sample of manganese-52 and measures its

LUCKY_DIMON [66]

Answer: 11.2 days

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.693}{5.6days}=0.124days^{-1}$

b) for completion of 75% of reaction

$t=\frac{2.303}{0.124}\log\frac{100}{100-75}$

$t=\frac{2.303}{0.124}\log\frac{100}{25}$

$t=11.2days$

The time for which she has to perform the experiment is 11.2 days

5 0

2 years ago

Assessment started: undefined.

Vilka [71]

B) the bacteria that live in the intestine of a rabbit

6 0

3 years ago

A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre

Goryan [66]

Answer:

$\boxed {\boxed {\sf 82.7 \textdegree C}}$

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

$\frac {P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}$

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

$\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58 \ atm *2.46 \ m^3}{T_2}$

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

$(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)$

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

$\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

$T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}$

The units of atmospheres and cubic meters cancel.

$T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}$

Solve inside the parentheses.

$T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}$

$T_2= \frac{369.246}{4.4625} \textdegree C}$

$T_2 = 82.74420168 \textdegree C$

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

$T_2 \approx 82.7 \textdegree C$

The temperature is approximately <u>82.7 degrees Celsius.</u>

3 0

2 years ago

In Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed

finlep [7]

The ratio of H⁺ ions to OH⁻ ions at a pH = 2 is 10¹⁰

<h3>Further explanation</h3>

Given

ph = 2

Required

The concentration of H⁺ and OH⁻ ions

Solution

The concentration of H⁺ ions

pH=-log[H⁺]

2=-log[H⁺]

[H⁺]=10⁻²

The concentration of OH⁻ ions

pH+pOH=14

pOH=14-2

pOH=12

pOH=-log[OH⁻]

12=-log[OH⁻]

[OH⁻]=10⁻¹²

The ratio of H⁺ ions to OH⁻ ions at a pH = 2

$\tt \dfrac{10^{-2}}{10^{-12}}=10^{10}$

6 0

2 years ago

Oxidation number of Mg3N2

Fudgin [204]

Answer:

n=-3

Explanation:

Mg = +2

3(+2) + 2N = 0

N = -3

8 0

3 years ago