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2 years ago

How many moles of MgS2O3 are in 201g of the compound?

Chemistry

1 answer:

Tamiku [17]2 years ago

8 0

<span>divide the 201g by the mol mass of the compound. Just add up the masses of the various element</span>

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At higher elevations, the boiling point of water decreases, due to the decrease in atmospheric pressure. As a result, what concl

White raven [17]

I believe the correct response is A. At higher elevations it would take less time to hard boil an egg, because there is less atmospheric pressure.

5 0

2 years ago

What is the pressure inside a 750 mL can of deodorant that starts at 15 degrees Celsius and 1.0 atm if the temperature is raised

Sonbull [250]

The answer is: the pressure inside a can of deodorant is 1.28 atm.

Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.

p₁ = 1.0 atm.; initial pressure

T₁ = 15°C = 288.15 K; initial temperature.

T₂ = 95°C = 368.15 K, final temperature

p₂ = ?; final presure.

1.0 atm/288.15 K = p₂/368.15 K.

1.0 atm · 368.15 K = 288.15 K · p₂.

p₂ = 368.15 atm·K ÷ 288.15 K.

p₂ = 1.28 atm.

As the temperature goes up, the pressure also goes up and vice-versa.

6 0

2 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

2 years ago

A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask ab

77julia77 [94]

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

By definition, $pH = -log[H_3O^+]$ .
NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$ .
From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$ .
Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$ .
We also know that $pOH = 14.00 - pH = -log[NaOH]$ . Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$ .
Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$ .

Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:

$m_{NaOH} = 10^{12.00 - 14.00}\cdot 39.997 g/mol\cdot 0.100 L = 0.0400 g$

7 0

2 years ago

Calculate the standard potential, e°, for this reaction from its equilibrium constant at 298 k. 6.47 x 10^5

Elina [12.6K]

The working equation for this is written below:

E° = 0.0592logK/n

So, we have to know the value of n first which represents the number of moles electron in the reaction. We cannot answer this because we are not given with the reaction. However, just suppose the reaction is:

Cu⁺ + 2e⁻ --> Cu

Then, n=2. Continuing,

E° = 0.0592log(6.47×10⁵)/2 =<em> 0.172 V</em>

4 0

2 years ago