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dedylja [7]
3 years ago
12

Srihari has a torch with him and 2 new batteries. However, the torch requires 3 batteries to work. Is there any way Srihari can

get the torch to glow?
Chemistry
1 answer:
zloy xaker [14]3 years ago
6 0

Answer:

go to the store to buy more batteries

Explanation:

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In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq
mariarad [96]
Assuming that the reactants are:

(NH4)2SO4 (aq) + Ba(NO3)2 (aq)

and the products are:

BaSO4 (s) + 2NH4NO3 (aq),

then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.

According to the solubility rules, the following elements are considered insoluble when paired with SO4:

Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+

Therefore, the precipitate will be BaSO4 (s).

5 0
3 years ago
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Which of the following is not an example of a molecule?
9966 [12]
D. F

Molecules are a group of bonded atoms but Fluorine stands on its own
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3 years ago
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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
adell [148]

Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)

we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

\Delta H'=6\times \Delta H_1+\Delta H_2

\Delta H'=6\times 414.0 kJ+132.0kJ

\Delta H'=2,626 kJ

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

7 0
3 years ago
. Element X has a nucleon (mass) number of 19 and a proton (atomic) number of 9.
Marina86 [1]

Answer:

C

Explanation:

it belong to that group as it needs 1 electron to be chemically stable

7 0
3 years ago
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
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