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belka [17]
3 years ago
14

How many moles of methane occupy a volume of 2.00 l at 50.0°c and 0.500 atm answer?

Chemistry
1 answer:
Sauron [17]3 years ago
5 0
Data Given:
                  Pressure  =  P  =  0.5 atm
 
                  Volume  =  V  =  2.0 L

                  Temperature  =  T  =  50 °C + 273  =  323 K

                  Moles  =  n  =  ?

Solution:
              Let suppose the gas is acting Ideally, Then According to Ideal Gas Equation.

                              P V  =  n R T
Solving for n,
                              n  =  P V / R T

Putting Values,
                             n  =  (0.5 atm × 2.0 L) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 323 K)

                             n  =  0.0377 mol
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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

8 0
3 years ago
A weighed piece of magnesium ribbon is added to a dried crucible, which is reweighed and heated in air to form the compound MgO.
olga nikolaevna [1]

Answer:

One of the errors for low percentage of magnesium could be because not all the magnesium may have reacted.

Explanation:

During the heating process, if the magnesium have not reacted completely, it can lead to low percentage of magnesium in the oxide formed. The product may still look a bit greyish rather than whitish after the heating process.

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3 years ago
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t
Sloan [31]

Answer:

Mass =  42.8g

Explanation:

4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).

Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol

Oxygen = 63.4g × 1mol / 32g = 1.9813mol

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.

Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.

5 moles of O2  = 6 moles of H2O

1.9831 moles = x

x = (1.9831 * 6 ) / 5

x = 2.37972 moles

Mass of H2O = Molar mass * Molar mass

Mass = 2.7972 * 18

Mass =  42.8g

6 0
3 years ago
Characteristics of an atom
Doss [256]

Answer:

It is composed of protons, which have a positive charge, and neutrons, which have no charge. Protons, neutrons, and the electrons surrounding them are long-lived particles present in all ordinary, naturally occurring atoms. Other subatomic particles may be found in association with these three types of particles.

Explanation:

Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged).

3 0
3 years ago
A graduated cylinder is filled to the 40.00-mL mark with a mineral oil. The masses of the cylinder before and after the addition
castortr0y [4]

Answer:

Firstly, Let's experiment this !

Experiment 1 :

159.446g - 124.966g = 34.48g

34.48g = The mass of Mineral oil.

The density of the mineral oil = M/V = 34.48g/40mL = 0.862g/cm³.

Experiment 2 :

124.966 + 18.173 = 143.139 = The mass of solid + cylinder.

124.966 + 50.952 = 175.918 = The mass of solid + cylinder + Mineral water.

175.918 - 143.139 = 32.779 = The mass of added mineral oil.

Explanation:

Now we have to find the volume of the added mineral oil using the density from experiment 1.

V = 32.779g/0.862g/cm³ = 38.02668213mL

Since we found the volume of the solid, we then have to subtract the added mineral oil volume from the total volume from experiment 1.

Volume of solid = 40-38.02668213 = 1.97331787mL

Density of solid = 18.713g/1.97331787mL = 9.483013499g/cm^3

1.97331787 = (4/3)(3.14)r³

1.97331787*(3/4)(3.14) = .4713338861

.4713338861 = r ³

r = 0.7782328425158433

r = 0.78

Now that's our final answer ! r = 0.78

5 0
3 years ago
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