All categories

3 years ago

29.2 grams of methane (carbon tetrahydride), CH4, is how

Chemistry

1 answer:

Elena-2011 [213]3 years ago

5 0

Answer:

1.83moles

Explanation:

Given parameters:

Mass of methane given = 29.2g

Unknown:

Number of moles = ?

Solution:

To find the number of moles in this mass of a compound;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of CH₄ = 12 + 4(1) = 16g/mol

Now;

Number of moles = $\frac{29.2}{16}$ = 1.83moles

You might be interested in

What volume is occupied by 35.2 g of carbon tetrachloride if its density is 1.60 g/ml?

ycow [4]

Density is mass divided by volume. Therefore, volume is mass divided by density.

$V=\frac{m}{\rho}=\frac{35.2g}{1.60\frac{g}{ml}}=22.0\ ml$

8 0

3 years ago

Read 2 more answers

If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?

Greeley [361]

Answer:

1.44 L

Explanation:

Since 25 is constant it is no use. Now, rearrange the gas formula. You should get...

P1V1/T2=P2V2T1

Next, rearrange to fit the problem. You should get...

V2=P1V1/P2

Fill in our values and solve. You should get 1.44 L

We can check this by knowing that P and V at constant T have an inverse relationship. Hence, this is correct.

- Hope that helps! Please let me know if you need further explanation.

8 0

3 years ago

How many grams of nickel (+2) are required to replace all of the silver when 15.55 grams of AgNO3 are present?

Karolina [17]

2AgNO3 + Ni2+ = Ni(NO3)2 + 2Ag<span>+</span>

From the reaction, it can be seen that AgNO3 and Ni2+ has following amount of substance relationshep:

n(AgNO3):n(Ni)=2:1

From the relationshep we can determinate requred moles of Ni2+:

n(AgNO3)=m/M= 15.5/169.87=0.09 moles

So, n (Ni)=n(AgNO3)/2=0.045 moles

Finaly needed mass of Ni2+ is:

m(Ni2+)=nxM=0,045x58.7=2.64g

8 0

3 years ago

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

3 0

3 years ago

A student was asked to prepare 500.0 mL of 6.0 M NaOH. The student measured 120.0 g of NaOH and placed it in a 1000 mL beaker. T

Morgarella [4.7K]

Answer:

The concentration of the solution will be much lower than 6M

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

n= CV

n = number of moles m/M( m= mass of solid, M= molar mass of compound)

C= concentration of substance

V= volume of solution

m=120g

M= 40gmol-1

V=500ml

120/40= C×500/1000

C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

7 0

3 years ago