7. A change of state is a(n)_process.
a. Irreversible
b. Reversible

Answers

Answer 1
Answer:

Answer: b

A change of state is a reversible process.

Answer 2
Answer: B that’s the answer

Good luck

Related Questions

A potassium ion (K+) would most likely bond with _____.Mg+OCl-Na+
A certain first-order reaction has a rate constant of 2.75 10-2 s−1 at 20.°c. what is the value of k at 45°c if ea = 75.5 kj/mol? webassign will check your answer for the correct number of significant figures. 0.0352 incorrect: your answer is incorrect.
Identify two ions that have the following ground-state electron configurations Part B[Ar]3d^5Check all that apply.A- Fe2+B- Fe3+C- Mn2+D- V+E- Sc2+
A chemical equation is shown:Al + O2 A1203According to the law of conservation of mass, how many atoms of oxygen exist in the products of this reaction?6532
What is the initial temperature of a gas if the volume changed from 1.00l to 1.10l and the final temperature was determined to be 255?

How much is a share in namsek worth as a percent of a share in oxd group

Answers

Answer:

10% is a share in NAMSEK worth as a percentage of a share in ODX Group Inc., in year 4

EXPLANATION:give brainliest

The share of NAMSEK in year 4 is around $  shares (as given from the above table). Hence the value of one share will be $2.5/share

The share of ODX Group Inc., in year 4 is around $ shares (as given from the above table). Hence the value of one share is $

Therefore a share in NAMSEK worth as a percentage of a share in ODX Group Inc., in year 4 will be  

= 0.1 or 10 %

Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water: Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

Answers

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

n = (m)/(M)

Where:

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

n = (m)/(M) = (14 g)/(56.1056 g/mol) = 0.249 mol

The enthalpy change is:

\Delta H = -43 (kJ)/(mol)*0.249 mol = -10.71 kJ

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

To create an image, the lens of an eye ________focuses light on the cornea
reflects light away from the retina
bends the cornea to correct vision
focuses light on the retina

Answers

It focuses light on the retina.

Answer:

focuses light on the retina

Explanation:

QuestionArrange the elements according to atomic radius.
Largest radius to Smallest radius
Answer Bank
K
Ca
Ga
Ge
As
Sc
Br
Kr

Answers

The elements according to the decreasing atomic radius are arranged as-

K, Ca, Sc, Ga, Ge, As, Br, Kr

An atomic radius is half the distance between adjacent atoms of the same element in a molecule. It is a measure of the size of the element’s atoms, which is typically the mean distance from the nucleus centre to the boundary of its surrounding shells of the electrons.

An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. The atomic radius decreases on moving from left to right across a period.

Thus the elements according to the decreasing atomic radius are arranged as -

K, Ca, Sc, Ga, Ge, As, Br, Kr

Learn more about Atomic radius, here:

brainly.com/question/13963619

#SPJ6

Largest

K
Ca
Ga
Ge
As
Br
Kr
Smallest

What is the electron configuration for26 / 12 Mg^+2

1s2 2s2 2p6 3s2 3p2


1s2 2s2 2p6 3s2 3p6 4s2 3d4


1s2 2s2 2p6


1s2 2s2 2p6 3s2 3p6 4s2 3d6


1s2 2s2 2p6 3s2

Answers

Answer:

The number of electrons for the Mg atom are 12 electrons. The electron configuration of magnesium is,

Mg (Z= 12) = 1s2 2s2 2p6 3s2

Explanation:

The first two electrons is placed in the 1s orbital. The 1s orbital can accommodate two electrons.

The next 2 electrons for magnesium go in the 2s orbital.

The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons.

We’ll put six in the 2p orbital and then put the remaining two electrons in the 3s.

Therefore, the Magnesium electron configuration will be 1s22s22p63s2.

PLS MARK ME AS BRAINLIEST

A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answers

Answer: The molal boiling point elevation constant k_b of X is 2.4^0C/m

Explanation:

Formula used for Elevation in boiling point :

\Delta T_b=k_b* m

or,

T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* w_1)

where,

T_b-T^o_b =(125.2-120.7)^0C=4.5^0C

k_b = boiling point constant  = ?

m = molality

w_2 = mass of solute (urea) = 55.4 g

w_1 = mass of solvent  X =  500 g

M_2 = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

4.5^oC=k_b* (55.4g* 1000)/(60* 500g)

k_b=2.4^0C/m

Thus the molal boiling point elevation constant k_b of X is 2.4^0C/m