I need help with electron affinity chemistry problem. I’m having a tough time.

If you use a horizontal force of 32.0 N to slide a 12.5 kg wooden crate across a floor at a constant velocity, what is the coeff

sergejj [24]

Answer:

Value of coefficient of kinetic friction is 0.26 .

Explanation:

Given:

Mass of wooden crate, m = 12.5 kg.

Horizontal force to keep the block moving with constant velocity, F = 32.0 N.

Since, the block is moving with constant velocity.

So, net force experience by it is zero.

Therefore, fore of friction is equal to applied force.

Now, force of friction , $F=\mu_kN$ ( here $\mu_k$ is coefficient of kinetic friction and N is normal force)

Therefore, $\mu_kN=\mu_k\times mg=\mu_k\times 12.5 \times 9.8=122.5\times \mu_k$

Now, both forces are equal.

$122.5\times \mu_k=32\\\mu_k=\dfrac{32}{122.5}=0.26$

The value of coefficient of kinetic friction is 0.26 .

Hence, it is the required solution.

6 0

3 years ago

Isotopes with unstable nuclei are_______and are called _______.

Inga [223]

Answer:

Radioactive and are often called radioisotopes

8 0

3 years ago

How do you balance Na + MgF{2} Na + Mg?

cricket20 [7]

2Na+MGF2=2NAF+MG
I hope that's right :)

6 0

3 years ago

What is the nuclear binding energy of an atom that has a mass defect of 5.0446

notsponge [240]

Answer:

Option C: 4.54 x $10^{11}$ KJ/mol of nuclei

Note: Here in this question option C is not correctly put. It is 4.54 x $10^{11}$ rather than 4.54 x $10^{-123}$ .

Explanation:

If mass defect is known, then nuclear binding energy can easily be calculated, here's how:

First step is to convert that mass defect into kg.

Mass defect = 5.0446 amu

Mass defect = 5.0466 x 1.6606 x $10^{-27}$

Because 1 amu = 1.6606 x $10^{-27}$ Kg.

Mass defect = 8.383 x $10^{-27}$ kg.

Now, we need to find out it's energy equivalent by using following equation:

Using the equation E = mc²:

where c= 3.00 x $10^{8}$ m/s²

E = (8.383 x $10^{-27}$ ) x (3.00 x $10^{8}$ )²

E = 7.54 x $10^{-10}$ J this energy is in Joules but nuclear binding energy is usually expressed in KJ/mol of nuclei. Let's convert it:

(7.54 x $10^{-10}$ Joule/nucleus)x(1 kJ/1000 Joule)x(6.022 x $10^{23}$ nuclei/mol) =

4.54 x $10^{11}$ kJ/mol of nuclei .

E = 4.54 x $10^{11}$ kJ/mol of nuclei . So, this is the nuclear binding energy of that atom, which is option C.

Note: Here in this question option C is not correctly put. It is 4.54 x $10^{11}$ rather than 4.54 x $10^{-123}$

4 0

3 years ago

Read 2 more answers

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol

Kryger [21]

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant $K_c= \dfrac{[CO][H_2]}{[H_2O]}$

The equilibrium constant $K_c= \dfrac{(0.17 )(0.17)}{0.74}$

The equilibrium constant $K_c= 0.03905$

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

$H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17$

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

$K_c= \dfrac{[CO][H_2]}{[H_2O]}$

$0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}$

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265 or x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

3 0

3 years ago