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3 years ago

What is the volume of 1 lb of mercury. the density of mercury is 13.546g/ml?

1 answer:

7 0

<span>Take the inversion of density: 1mL/13.6 g and multiply it by the conversion factor 453.6 g/ 1 lb and the given 5.00 lb. Units for mass (grams) and units for weight (lbs) cancel leaving only units of volume. I believe it should be 167 mL or 0.167 L</span>

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What is liquor ammonia fortis?

erik [133]

Answer:

Ammonia fortis liquor is a saturated solution of ammonia in water. It is also called 880 ammonia. Its relative density is 0.880. It is stored in tightly sealed bottles in a cold place. (Sorry if I'm wrong)

Explanation:

6 0

3 years ago

Read 2 more answers

If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol

igomit [66]

Answer:

$\Delta _RH=4x10^{-3}\frac{kJ}{mol}$

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

$\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}$

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

$\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}$

Best regards.

5 0

3 years ago

4. How many weeks are equal to 2.94 x 10 to the power of 6 minutes?

OverLord2011 [107]

Answer:

292 weeks

Explanation:

5 0

3 years ago

When 120 g of carbon reacts completely with 320 g of oxygen the mass of carbon dioxide formed will be?

olga2289 [7]

440 cause mass cant be created or destroyed

6 0

3 years ago

Write the balanced standard combustion reaction for the c5h7on3s2

AlexFokin [52]

Answer:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Explanation:

When C5H7ON3S2 under go Combustion, the following are obtained as illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

Now, let us balance the equation. This is illustrated below:

C5H7ON3S2 + O2 —> CO2 + H2O + N2 + S2

There are 3 atoms of N on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of C5H7ON3S2 and 6 in front of N2 as shown below:

4C5H7ON3S2 + O2 —> CO2 + H2O + 6N2 + S2

There are 20 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 20 in front of CO2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + H2O + 6N2 + S2

There are 28 atoms on H on the left side and 2 atoms on the right side. It can be balance by putting 14 in front of H2O as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + S2

There are 8 atoms of S on the left side and 2 atoms on the right side. It can be balance by putting 4 in front of S2 as shown below:

4C5H7ON3S2 + O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, there are a total of 54 atoms of O on the right side and 6 atoms on the left side

It can be balance by putting 25 in front of O2 as shown below:

4C5H7ON3S2 + 25O2 —> 20CO2 + 14H2O + 6N2 + 4S2

Now, we can see that the equation is balanced.

4 0

3 years ago