Determine the mass in grams of 0.500 mol of uranium(molar mass =238.05g)

Problem PageQuestion A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of

igomit [66]

Answer:

5.37 × 10⁻⁴ mol/L

Explanation:

<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>

Step 1: Given data

Initial concentration (C₁): 0.00154 mol/L

Initial volume (V₁): 230. mL

Final concentration (C₂): ?

Final volume (V₂): 660. mL

Step 2: Calculate the concentration of the final solution

We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L

5 0

3 years ago

A 5.00 L sample of air at 0 C is warmed to 100.0 C. What is the new volume of the air? First, identify V1.

Paladinen [302]

Answer : The new volume of the air is, 6.83 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=5.00L\\T_1=0^oC=(0+273)K=273K\\V_2=?\\T_2=100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$\frac{5.00L}{273K}=\frac{V_2}{373K}\\\\V_2=6.83L$

Therefore, the new volume of the air is, 6.83 L

6 0

4 years ago

What numbers are shown in the symbol for a radioactive nuclide

Bingel [31]

The atomic # and the mass #.

6 0

3 years ago

Read 2 more answers

What mass of natural gas (ch4) must you burn to emit 272 kj of heat?

balandron [24]

CH4 + 2 O2 ---> CO2 + 2 H2O Q = 891,6 kJ / mol CH4

1 mol CH4 = 16 g

16 g ---- 891,6 kJ
x g ----- 272 kJ

x = 272 kJ × 16 g / 891,6 kJ = 4,88 g

You must burn 4,88 g of CH4.

:-) ;-)

7 0

3 years ago

Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide

Luda [366]

A. The balanced equation for the reaction is

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

B. The volume of the oxygen gas required to burn 0.700 L of acetylene gas is 1.75 L

C. The volume of carbon dioxide gas produced is 1.4 L

D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

Therefore,

0.7 L of C₂H₂ will react with = (0.7 × 5) / 2 = 1.75 L of O₂

Thus, 1.75 L of O₂ is needed for the reaction

<h3>C. How to determine the volume of carbon dioxide </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 4 L of CO₂

Therefore,

0.7 L of C₂H₂ will react to produce = (0.7 × 4) / 2 = 1.4 L of CO₂

Thus, 1.4 L of CO₂ were produced

<h3>D. How to determine the volume of water. </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 2 L of H₂O

Therefore,

0.7 L of C₂H₂ will also react to produce 0.7 L of H₂O

Thus, 0.7 L of H₂O was produced

Learn more about stoichiometry:

brainly.com/question/14735801

8 0

3 years ago

Determine the mass in grams of 0.500 mol of uranium(molar mass =238.05g)​

Determine the mass in grams of 0.500 mol of uranium(molar mass =238.05g)