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Sati [7]
3 years ago
8

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Chemistry
1 answer:
Lisa [10]3 years ago
4 0
The definition of potential energy is the energy that is absorbed within an object. The object is holding but not using the energy. In contrast, kinetic energy is the energy of motion so whenever something is moving it has kinetic energy.

The most classic example of going from potential to kinetic energy is the idea of a roller coaster. Once the roller coaster is going up towards a hill, it’s building up in potential energy. Once the coaster comes to the top of the hill, it contains total potential energy. And while the coaster begins to come down and pick up speed, this potential energy is being transferred as kinetic energy.

Hope that helps! :)
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How could you safely determine if a base is stronger than an acid? Compare the taste of the base and the acid. Use a conductivit
Gnoma [55]
Use blue litmus paper. This is an indicator that can safely determine whether it is a base or an acid by changing color in response to the substance. This color indicates whether it is an acid or a base. Refer to the pH scale to see if the substance is basic or acidic.
6 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
nguyên tử sắt có điện tích hạt nhân là 26+. trong nguyên tử, số hạt mang điện nhiều hơn số hạt không mang điện là 22. Hãy xác đị
dimaraw [331]

Answer:

I don't know yr language

8 0
3 years ago
A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t
erik [133]

Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

Answer:

-314 kJ

+628 kJ

+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

This reaction is the inverted reaction given multiplied by 4, so

ΔH = 4*(157) = +628 kJ

3) FeCl2(s) + H2(g) --> Fe(s) + 2HCl

This reaction is the inverted reaction given, so

ΔH = +157 kJ

4 0
3 years ago
The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles

b) moles of C_2H_4

\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

7 0
3 years ago
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