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Paraphin [41]
2 years ago
6

Identify the change of state occurring in each situation. your friend's perfume producing an aroma that fills the room making ic

e cream from milk in an ice cream maker formation of frost on a cold windshield overnight wax dripping from a candle naphthalene solid (found in moth balls) producing fumes in the closet to deter moths your breath fogging up on your car windshield on a cold winter day
Chemistry
1 answer:
Darina [25.2K]2 years ago
4 0

Answer:

The solution to this question can be defined as follows:

Explanation:

  • W_{ax} Naphthalene falls off from candle something, strong fumes-sublimation produces.
  • On a rainy day in winter, the breath fogging up on your cor's windshield.
  • Brost formation on an overnight deposition on a frozen windshield.
  • Making milk ice cream freezing the perfuming scent that fills the creation of the room.
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recheck Fusion reactions power the in space. Humans currently use fusion in , but researchers are seeking more applications.
quester [9]

One: looks to be correct for both answers. Certainly the first one is. The second depends on your other choices. But military use is one.

Two: is correct. Pd has (in this case) an atomic mass of 114 and its number is 46

Three: Even with my slop numbers, 4.98  is the answer (although I get 4.99 but again, my numbers are pretty sloppy).

Four: Slop numbers say 78.3, but 78 is the right answer.

Five: Slop numbers agree with Al2S3. I think that's D

They are all correct. Very Fine Work.

4 0
2 years ago
In the synthesis of barium carbonate from an alkali metal carbonate (M2CO3 where M is one of the alkali metals) a student genera
Paraphin [41]

Answer:

0.019 moles of M2CO3

Explanation:

M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)

From the equation above;

1 mol of  M2CO3 reacts to produce 1 mol of BaCO3

Mass of BaCO3 formed = 3.7g

Molar mass of BaCO3 = 197.34g/mol

Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187  ≈ 0.019mol

Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,

1 = 1

x = 0.019

x = 0.019 moles of M2CO3

3 0
2 years ago
An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g
jekas [21]

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

For example, O_{2}, N_{2} etc are pure substances.

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

  • pure chemical substance.
  • heterogenous mixture.
  • mixture.
4 0
2 years ago
(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path
never [62]
<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>
4 0
2 years ago
Read 2 more answers
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
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