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choli [55]
2 years ago
13

A student sets up the following equation to convert a measurement.

Chemistry
1 answer:
sleet_krkn [62]2 years ago
6 0

Answer:

\frac{1  kJ}{1,000 J}  ·  \frac{1,000 mmol}{1 mol}

Explanation:

The final answer has a different set of units. In particular, millimoles (mmol) changes to moles (mol) and joules (J) changes to kilojoules (kJ). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since J is located in the numerator, it must be located in the denominator of the conversion. Furthermore, since mmol is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportions:</u>

1,000 J = 1 kJ

1,000 mmol = 1 mol

<u>The full expression:</u>

<h2>-1.3*10^4\frac{J}{mmol*^oC}  ·  \frac{1  kJ}{1,000 J}  ·  \frac{1,000 mmol}{1 mol}  =  ? \frac{kJ}{mol*^oC}
                        ^           ^</h2>

As you can see, the old units cancel out and you are left with kJ in the numerator and mol in the denominator.

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3 years ago
Reduce the following equation to a net ionic equation: 2Na+ + SO4^2- + Ba^2+ + 2Cl^- ---&gt; 2Na^+ + 2Cl^- + BaSOv4
Eduardwww [97]

net ionic equation

B) SO₄²⁻ (aq) + Ba²⁺ (aq) → BaSO₄ (s)

Explanation:

We have the following chemical equation:

2 Na⁺ (aq) + SO₄²⁻ (aq) + Ba²⁺ (aq) + 2 Cl⁻ (aq) → 2 Na⁺ (aq) +  2 Cl⁻ (aq) + BaSO₄ (s)

To get the net ionic equation we remove the spectator ions and we get:

SO₄²⁻ (aq) + Ba²⁺ (aq) → BaSO₄ (s)

were:

(aq) - aqueous

(s) - solid

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net ionic equation

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3 years ago
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You are given two aqueous solutions with different ionic solutes (Solution A and Solution B). What if you are told that Solution
klio [65]

Answer:

Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.  

It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,  

% by mass = mass of solute/mass of solution * 100

Now the formula for molality is,  

Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams

Now molality of solution A is,  

m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)

m = 2.07

Now the molality of solution B is,  

m = 15/58.5 * 1000/85

m = 3.02

Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).

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3 years ago
Which of the following is an example of a way you can reduce personal consumption of resources?
dsp73

Answer:

B. using efficient light bulbs

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3 years ago
Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2
Mariana [72]

<u>Answer:</u> The value of \Delta S^o for the reaction is 1051.93 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K

Hence, the value of \Delta S^o for the reaction is 1051.93 J/K

5 0
4 years ago
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