All categories

3 years ago

What is the mass of a cube measuring 3cm on each side that had a density of 8.1g/cm3?

Chemistry

1 answer:

Hunter-Best [27]3 years ago

6 0

Answer:

48.6 g

Explanation:

Given data:

Length of one side = 3 cm

Density of cube = 8.1 g/cm³

Mass of cube = ?

Solution:

First of all we will calculate the volume of cube.

V = length × height × width

V = 3 cm × 3cm× 3cm

V = 27cm³

Now we will determine the mass by using density formula.

d = m/v

d = density

m = mass

v = volume

8.1 g/cm³ = m/ 27 cm³

m = 8.1 g/cm³ ×27 cm³

m = 48.6 g

You might be interested in

A container with a volume of 893L contains how may moles of air at STP?

Vadim26 [7]

Answer:

i think 54

Explanation:

7 0

3 years ago

Anyone (cool, funny, middle school) wanna start a conversation?

pychu [463]

Answer:

sure

Explanation:

what's up

5 0

3 years ago

Read 2 more answers

Which of the following is a cause of a dna mutation

lisov135 [29]

Answer:

Can you please tell us what the following are?

5 0

3 years ago

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

Answer: The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial: 0.275

At eqllm: 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 218 kJ/mol and the change in enthalpy for the

madreJ [45]

Answer:

$470\frac{KJ}{mol}$

Explanation:

The activation energy represents the energy barrier that reagents must pass to transform into products (or products to transform into reagents in a reverse reaction)

For any reaction, the change in enthalpy is related to the activation energy by the equation

$\Delta H =E_{a\ direct}-E_{a\ reverse}$

So, the activation energy for the reverse reaction is

$E_{a\ reverse}=E_{a\ direct}-\Delta H =218 \frac{KJ}{mol} - (-252)\frac{KJ}{mol}=470\frac{KJ}{mol}$

4 0

3 years ago