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3 years ago

In a controlled experiment, the single factor manipulated by the researcher is

Chemistry

1 answer:

Georgia [21]3 years ago

6 0

Answer:

False

Explanation:

It is the opposite. A good way to remember is the that the dependent variable depends on the independent variable.

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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-

fomenos

Answer:

The total pressure after one half is 6.375 atm.

Explanation:

The initial pressure of product is increases while the pressure of reactant would decrease.

Balanced chemical equation:

2N₂O → 2N₂ + O₂

The pressure of N₂O is 5.10 atm. The change in pressure would be,

N₂O = -2x

N₂ = +2x

O₂ = +x

The total pressure will be

P(total) = P(N₂O) + P(N₂) + P(O₂)

P(total) = ( 5.10 - 2x) + (2x) + (x)

P(total) = 5.10 + x

After one half life:

P(N₂O) = 1/2(5.10) = 5.10 - 2x

x = 5.10 - 1/2(5.10) /2

x = 5.10 - 0.5 (5.10) /2

x = 5.10 - 2.55 / 2

x = 2.55 /2 = 1.275 atm

Thus the total pressure will be,

P(total) = 5.10 + x

P(total) = 5.10 + 1.275

P(total) = 6.375 atm

4 0

3 years ago

How many elements are there

yawa3891 [41]

There are 118 known elements

4 0

4 years ago

Compute the molar enthalpy of combustion of glucose (C6 H12O6 ): C6 H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2 O (g) Given that com

lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30 = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0

3 years ago

Which example is an endothermic process

Annette [7]

Answer:

Endothermic reactions is the result of heat being absorbed.

Explanation:

Melting ice cubes. Melting solid salts. Evaporating liquid water. Converting frost to water vapor (melting, boiling, and evaporation, in general, are endothermic processes.

(These are all examples of endothermic reactions.)

4 0

3 years ago

Which of the following is TRUE? A) The equivalence point is where the amount of acid equals the amount of base during any acid-b

liberstina [14]

Answer:

the correct option is B

Explanation:

The correct option is b, since if we reach pH 7, it means that the acid-base reaction is neutralized, therefore the base has been neutralized by an acid or vice versa, without taking into account the proteins or the amounts of both components .

5 0

3 years ago

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