All categories

3 years ago

What would happen if you left your dichloromethane/ethanol solvent mixtures uncovered?

1 answer:

3 0

Dichloromethane/ethanol solvent mixtures are substances composed of compounds that has a very high volatility. The higher the volatility of the substances, the faster it is for them to vaporized.

If we leave the solution or the mixture uncovered then, it is easier for the vapors produced to escape the system to the atmosphere.

You might be interested in

A solution has a [OH−] of 1 x 10−12. What is the pOH of this solution?<br> 2<br> 7<br> 10<br> 12

strojnjashka [21]

-log (1×10^-12) is how you calculate the pOH which in this case is 12

7 0

3 years ago

Read 2 more answers

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel ro

melamori03 [73]

<u>Answer:</u> The mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For nickel:</u>

Given mass of nickel = 14.8 g

Molar mass of nickel = 58.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of nickel}=\frac{14.8g}{58.7g/mol}=0.252mol$

For the given chemical reaction:

$3NiO(s)+2Al(s)\rightarrow 3Ni(l)+Al_2O_3(s)$

<u>For nickel (II) oxide:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 3 moles of nickel (II) oxide

So, 0.252 moles of nickel will be produced from $\frac{3}{3}\times 0.252=0.252mol$ of nickel (II) oxide

Now, calculating the mass of nickel (II) oxide by using equation 1:

Molar mass of nickel (II) oxide = 74.7 g/mol

Moles of nickel (II) oxide = 0.252 moles

Putting values in equation 1, we get:

$0.252mol=\frac{\text{Mass of nickel (II) oxide}}{74.7g/mol}\\\\\text{Mass of nickel (II) oxide}=(0.252mol\times 74.7g/mol)=18.8g$

<u>For aluminium:</u>

By Stoichiometry of the reaction:

3 moles of nickel are produced from 2 moles of aluminium

So, 0.252 moles of nickel will be produced from $\frac{2}{3}\times 0.252=0.168mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1:

Molar mass of aluminium = 27 g/mol

Moles of aluminium = 0.168 moles

Putting values in equation 1, we get:

$0.168mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=(0.168mol\times 27g/mol)=4.54g$

Hence, the mass of nickel (II) oxide and aluminium that must be used is 18.8 g and 4.54 g respectively.

4 0

3 years ago

Helppp plsss !!!!!!!!!

geniusboy [140]

Answer:

B. because there is two equations just like commutative property in math its the same thingish

Explanation:

5 0

3 years ago

4 NH3 + 7 O2 → 4 NO2 + 6 H2O What is the mole ratio between oxygen and nitrogen dioxide? 7 moles to 6 moles 4 moles to 6 moles 7

Mashutka [201]

Answer:

7:4

Explanation:

O2 : NO2

7 : 4

hope this helps :)

8 0

2 years ago

Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.

Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>

Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
Initial temperature (T₁) = –25 °C
Final temperature (T₂) = 0 °
Change in temperature (ΔT) = 0 – (–38) = 38 °C
Specific heat capacity (C) = 2050 J/(kg·°C)
Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>

Mass (m) = 0.4 Kg
Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>

Mass (M) = 0.4 Kg
Initial temperature (T₁) = 0 °C
Final temperature (T₂) = 100 °C
Change in temperature (ΔT) = 100 – 0 = 100 °C
Specific heat capacity (C) = 4180 J/(kg·°C)
Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>

Mass (m) = 0.4 Kg
Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>

Mass (M) = 0.4 Kg
Initial temperature (T₁) = 100 °C
Final temperature (T₂) = 160 °C
Change in temperature (ΔT) = 160 – 100 = 60 °C
Specific heat capacity (C) = 1996 J/(kg·°C)
Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>

Heat for –38 °C to 0°C (Q₁) = 31160 J
Heat for melting (Q₂) = 133600 J
Heat for 0 °C to 100 °C (Q₃) = 167200 J
Heat for vaporization (Q₄) = 904000 J
Heat for 100 °C to 160 °C (Q₅) = 47904 J
Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0

2 years ago