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2 years ago

What are two variables that affect kinetic energy

Chemistry

1 answer:

zubka84 [21]2 years ago

7 0

The answer for the following questions is explained below.

Explanation:

The two variables that affect kinetic energy are:

mass and
velocity

velocity - The faster an object moves,the more the kinetic energy it has.

mass - Kinetic energy increases as mass increases

The kinetic energy of an object depends on both its mass and its velocity

Kinetic energy increases as mass increases

For example,think about rolling a bowling ball and a golf ball down a bowling lane at same velocity

Here,the bowling ball has more mass than the golf ball

Therefore you use more energy to roll the bowling ball than to roll the golf ball

The bowling ball is more likely to knock down the pins because it has more kinetic energy than the golf ball

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Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C

Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, $E = \triangle M * 931.5$

$\triangle M = \sum M_{product} - \sum M_{reactant}$

Mass of 1H, $M_{H} = 1.007823$

Mass of 2H, $M_{2H} = 2.0141u$

Mass of 3H, $M_{3H} = 3.016 u$

Mass of Helium, $M_{4He} = 4.002602u$

Mass of Beryllium, $M_{7Be} = 7.01693 u$

Mass of neutron, $M_{n} = 1.008664 u$

a) $2H + 3H \rightarrow 4He + n$

$\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u$

Energy released,

$E = -0.01884 * 931.5\\E = -17.55 Mev$

Energy released = 17.55 MeV

b) $4He + 4He \rightarrow 7Be + n$

$\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} + M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u$

Energy released,

$E = 0.02039 * 931.5\\E = 18.99 Mev$

c) $2H + 2H \rightarrow 3 He$ + n

$\triangle M = (M_{3He} + M_{n} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u$

Energy released,

$E = -0.003536 * 931.5\\E = -3.29 Mev$

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) $2H + 2H \rightarrow 3H + 1H$

$\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u$

$E = -0.00435 * 931.5\\E =-4.075 Mev$

E = 4.075 MeV ( Energy is released)

5 0

2 years ago

The basic fact that determines molecular shapes is that

Alinara [238K]

Answer:

The shape of a molecule is mostly determined by repulsion among the pairs of electrons around a central atom.

4 0

2 years ago

Compound formula: MgCl2

umka2103 [35]

Answer:

please where are the questions and also make your questions clear.

7 0

2 years ago

Read 2 more answers

Hello!

blondinia [14]

Answer:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.00064M$

Explanation:

Hello there!

In this case, according to the given chemical reaction at equilibrium, we can set up the equilibrium expression as follows:

$K=\frac{[CO][Cl_2]}{[COCl_2]}$

Which can be written in terms of x, according to the ICE table:

$0.32=\frac{x^2}{0.015M-x}$

Thus, we solve for x to obtain that it has a value of 0.01436 M and therefore, the concentrations at equilibrium turn out to be:

$[CO]=[Cl_2]=0.01436M$

$[COCl_2]=0.015M-0.01436M=0.00064M$

Regards!

5 0

2 years ago

What is the oxidation state of Hg in Hg2Cl2?

Anna [14]

Answer:

Explanation:

Electrochemistry. In oxidation–reduction (redox) reactions, electrons are transferred from one A redox reaction is balanced when the number of electrons lost by the reductant Hg(l)∣Hg2Cl2(s)∣Cl−(aq) ∥ Cd2+(aq)∣Cd(s).

As is evident from the Stock number, mercury has an oxidation state of +1. This makes sense, as chlorine usually has an oxidation state of -1.

5 0

2 years ago