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klio [65]
3 years ago
15

Explain the concept of dihyric alcohols​

Chemistry
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

An alcohol that contains two hydroxyl groups

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In reality, energy conversion from burning fuel is never 100% efficient. Significant loss of energy due to heating occurs. If th
algol [13]

Answer:

Electrical energy = 130000000 J and Heat energy = 520000000 J

Explanation:

Multiply the amount of joules from the last question (650000000) by .20 and .80. (Which are the percentages)

6 0
3 years ago
2NO(g) N2(g) + O2(g); Keq = 2.1 × 1030 What does the value of the equilibrium constant indicate?
olga_2 [115]
The equilibrium constant is a value which represents the equilibrium of a reaction. It is a reaction quotient when the reaction reached equilibrium. If Keq is greater than 1, the mixture contains mostly the products. On the other hand, if Keq is less than 1, the mixture contains the reactants. For this case, the mixture contains mostly products.
7 0
3 years ago
Name two elements that you would expect to have properties very much like those of calcium
Semmy [17]
Beryllium, magnesium, strontium, barium, or radium. Hope this helped :))
7 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
Experiments show that each of the following redox reactions is second-order overall: (1) NO2(g) + CO(g) → NO(g) + CO2(g) (2) NO(
MatroZZZ [7]

Answer:

a) rate law1 = k[NO2]²

b) rate law2 = k[NO][O3]

Explanation:

NO2(g) + CO(g) → NO(g) + CO2(g)

NO(g) + O3(g) → NO2(g) + O2(g)

When [NO2] in reaction 1 is doubled, the reaction quadruples

Rxn is second order.

rate law1= [NO2]^a [CO]^b

rate law1= [NO2]² [CO]^0

rate law1 = k[NO2]²

When [NO] in reaction 2 is doubled, the rate doubles.

Rxn is first order

The ratio is 1:1

this makes the rate law2 = k[NO][O3]

6 0
3 years ago
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