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Kisachek [45]
3 years ago
6

Why are chemists interested in studying thermochemistry? a. Heat is often absorbed during reactions.b. Heat is often released du

ring a chemical reaction, but never absorbed. c. Heat is often released during chemical reactions.d. Heat is often absorbed in a chemical reaction, but never released.
Chemistry
1 answer:
Wittaler [7]3 years ago
6 0

Answer:

a. Heat is often absorbed during reactions.

c. Heat is often released during chemical reactions.

Explanation:

Thermochemistry -

The study of thermochemistry involve the change in the amount of heat , during any physical or chemical process , is referred to as thermodynamics .

The focus of thermochemistry is on the changing amount of energy in the form of heat , on the system with respect to the surroundings .

The process like boiling , melting , sublimation , may require energy or releases energy , and hence are studied under thermochemistry .

Hence , from the given question ,

The correct options are -  a , c.

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Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 13.2 moles of FeO?
irina [24]

The number of moles of aluminium  that are needed to react completely  with 13.2 moles of FeO  is  8.8 moles



calculation


2Al + 3FeO → 3aFe +Al2O3

by use of of mole ratio of Al: FeO  from  equation above = 2:3 the moles of  Al is  therefore

= 13.2  x  2/3=8.8  moles  of Al

3 0
3 years ago
Blance equation __CaBr2 (aq) + ___Li3PO4(aq) → ___Ca3(PO4)2(s) + ____LiBr (aq)
QveST [7]

Answer:

3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr

Explanation:

The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.

Here was the process:

CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr

Balances PO4 (2on both sides)

CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Lithiums (6 on each side)

3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)

Hope this helped!

4 0
3 years ago
A saturated solution of pbcl2 is found to contain 2.88 x 10-2 m chloride ions. what is the correct value of the solubility produ
bazaltina [42]

The solubility equilibrium of PbCl_{2}:

PbCl_{2}(aq)  Pb^{2+}(aq) + 2 Cl^{-}(aq)

K_{sp}=[Pb^{2+}][Cl^{-}]^{2}

[Cl^{-}] = 2.88 * 10^{-2} M

[Pb^{2+}]=\frac{[Cl^{-}]}{2} = \frac{2.88 * 10^{-2}}{2}=1.44 *10^{-2}

K_{sp}=[Pb^{2+}][Cl^{-}]^{2}

= (1.44 * 10^{-2})(2.88*10^{-2})^{2}

= 1.19 * 10^{-5}

So, the corrected solubility product will be 1.19 * 10^{-5}

4 0
3 years ago
What does the word resources mean as it is used in the passage?
shutvik [7]
The answer to this will be A
6 0
3 years ago
Determine the rate law and the value of k for the following reaction using the data provided.CO(g) Cl2(g) → COCl2(g)[CO]i (M)[Cl
marshall27 [118]

<u>Answer:</u> The rate law expression is \text{Rate}=k[CO]^1[Cl_2]^{\frac{3}{2}} and value of 'k' is 11.04M^{-\frac{7}{2}}s^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

CO(g)+Cl_2(g)\rightarrow COCl_2(g)

Rate law expression for the reaction:

\text{Rate}=k[CO]^a[Cl_2]^b

where,

a = order with respect to carbon monoxide

b = order with respect to chlorine

  • Expression for rate law for first observation:

0.696=k(0.25)^a(0.40)^b       ....(1)

  • Expression for rate law for second observation:

1.970=k(0.25)^a(0.80)^b       ....(2)

  • Expression for rate law for third observation:

3.94=k(0.50)^a(0.80)^b       ....(3)

Dividing 2 from 3, we get:

\frac{3.94}{1.970}=\frac{(0.50)^a(0.80)^b}{(0.50)^a(0.80)^b}\\\\2=2^a\\a=1

Dividing 1 from 2, we get:

\frac{1.970}{0.696}=\frac{(0.25)^a(0.80)^b}{(0.25)^a(0.40)^b}\\\\2.83=2^b\\b=1.5

Thus, the rate law becomes:

\text{Rate}=k[CO]^1[Cl_2]^{\frac{3}{2}}

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

0.696=k[0.25]^1[0.40]^{\frac{3}{2}}\\\\k=11.04M^{-\frac{7}{2}}s^{-1}

Hence, the rate law expression is \text{Rate}=k[CO]^1[Cl_2]^{\frac{3}{2}} and value of 'k' is 11.04M^{-\frac{7}{2}}s^{-1}

6 0
3 years ago
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