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shutvik [7]
3 years ago
5

Which of the following equations can be used to determine the change in enthalpy of a system? A. `DeltaH_"reaction"=DeltaH_"prod

ucts"×DeltaH_"reactants"` B. `DeltaH_"reaction"=DeltaH_"products"÷DeltaH_"reactants"` C. `DeltaH_"reaction"= DeltaH_"products"+DeltaH_"reactants"` D. `DeltaH_"reaction" = DeltaH_"products"−DeltaH_"reactants"`
Chemistry
1 answer:
Nostrana [21]3 years ago
6 0

<u>Answer:</u> The correct answer is Option D.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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

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

Hence, the correct answer is Option D.

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Dmitri Mendeleev's Periodic Law states that chemical and physical properties repeat themselves in groups of _____. A. 4 B. 8 C.
krok68 [10]
<span>Dmitri Mendeleev's Periodic Law states that chemical and physical properties repeat themselves in groups of 8.</span>
6 0
3 years ago
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A gas that effuses through a porous cylinder 1.87 times faster than chlorine gas. what is the molar mass and identity.
Rufina [12.5K]
From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1 
mm1 = 71/1.87²
         = 71/3.4969
         = 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon
6 0
3 years ago
Mg(OH)2 + 2 HBr à MgBr2 + 2 H2O
AnnyKZ [126]

Explanation:

The balanced equation of the reaction is given as;

Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)

1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?

From the reaction;

2 mol of HBr produces 1 mol of  MgBr2

Converting to masses using;

Mass = Number of moles * Molar mass

Molar mass of HBr = 80.91 g/mol

Molar mass of MgBr2 = 184.113 g/mol

This means;

(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2

18.3g would produce x

161.82 = 184.113

18.3 = x

x = (184.113 * 18.3 ) / 161.82 = 20.8 g

2. How many moles of H2O will be produced from 18.3 grams of HBr?

Converting the mass to mol;

Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol

From the reaction;

2 mol of HBr produces 2 mol of H2O

0.226 mol would produce x

2 =2

0.226 = x

x = 0.226 * 2 / 2 = 0.226 mol

3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?

From the reaction;

2 mol of HBr reacts with 1 mol of Mg(OH)2

18.3g of HBr =  0.226 mol

2 = 1

0.226 = x

x = 0.226 * 1 /2

x = 0.113 mol

5 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

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3 years ago
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