What coefficients would balance the following equation?

Use enthalpies of formation to determine the ΔHreaction for the reaction

Daniel [21]

Answer:

The enthalpy is increased by the increased heat of the reaction.

Explanation:

In this reaction, as the transition from solid to liquid state, enthalpy increases, that is, the heat applied to change its state is greater and this increases, reaching a mayor disorder.

If the reaction increases its heat, and a liquid state passes, the enthalpy increases, therefore the disorder also and the entropy will also be increased.

5 0

3 years ago

When 55.0 grams of metal at 75.0°C is added to 100. grams of water at 15.0°C, the temperature of the water rises to 18.7°C. Assu

olga2289 [7]

Answer:

The specific heat of the metal is 0,50 J/gºC

Explanation:

Assume that no heat is lost to the surroundings

(Q = m . C . ΔT)metal + (Q = m . C . ΔT)water = 0

Let's replace our values.

55g . C . (18,7ºC - 75ºC) + 100g . 4,184 J/g·°C . (18,7ºC - 15ºC) = 0

55g . C . -56,3 ºC + 418,4J/·°C . 3,7ºC = 0

-3096,5 gºC . C + 1548,08 J = 0

1548,08 J = 3096,5 gºC . C

1548,08 J / 3096,5 gºC = C = 0,50 J/gºC

8 0

3 years ago

How many Liters of 0.50M HCl are needed to neutralize 0.050L of 0.101M Ba(OH)2?

Aleksandr-060686 [28]

Answer:

$V_{HCl}=5.05x10^{-3}L$

Explanation:

Hello,

In this case, since hydrochloric acid and barium hydroxide are in a 2:1 molar ratio, for the neutralization, the following moles equality must be obeyed:

$2*n_{HCl}=n_{Ba(OH)_2}$

In such a way, in terms of molarities and volumes, we can compute the required volume of hydrochloric acid as shown below:

$2*M_{HCl}V_{HCl}=M_{Ba(OH)_2}V_{Ba(OH)_2}\\\\V_{HCl}=\frac{M_{Ba(OH)_2}V_{Ba(OH)_2}}{2M_{HCl}} =\frac{0.101M*0.050L}{2*0.50M} \\\\V_{HCl}=5.05x10^{-3}L$

Besr regards.

8 0

3 years ago

Read 2 more answers

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$