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shtirl [24]
2 years ago
7

The products of a chemical reaction have different properties than the reactants.

Chemistry
1 answer:
Marta_Voda [28]2 years ago
5 0

Answer:

Explanation:

The products of a chemical reaction have different properties than the reactant because they underwent a chemical change by definition of a chemical reaction. The products are new combinations atoms forming different molecules.

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What is the concentration, in mass percent (m/m), of a solution prepared from 50.0 g nacl and 150.0 g of water?
Nataly [62]
Mass of solute ( m1 ) = 50.0 g

mass of solvent ( m2 ) = 150.0 g

Therefore:

m/m = ( m1 / m1 + m2 )

m/m = ( 50.0 / 50.0 + 150.0 )

m/m = ( 50.0 / 200 )

m/m = 0.25
8 0
3 years ago
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(( PLEASE HELP QUICK ))Which expression can be used to calculate the ratio of the rate of effusion of gas A to the rate of effus
borishaifa [10]

Answer:

The square root of the molar mass of B ÷ the square root of the molar mass of A  

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases A and B, the ratio of their rates of effusion is

\dfrac{r_{\text{A}}}{r_{\text{B}}} = \sqrt{\dfrac{M_{\text{B}}}{M_{\text{A}}}}

3 0
3 years ago
Suggest the identity of the gas in the bubbles ​
Vlad1618 [11]

Answer:

<h3>it is hydrogen gas </h3>

3. the chemical equation is :

MgSO4 ( s ) + H2SO4 (aq)-->

MgSO4 (aq) + CO2 (g) + H2O (I)

THE IONISING OF H2SO4 IN SOLUTION IS CONPLEX

4. MAGNESIUM REACTION WITH DILUTE SULPHURIC ACID TO FORM

MAGNESIUM SULPHATE AND HYDROGEN GAS

Explanation:

hope it helps u

7 0
3 years ago
Read 2 more answers
A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)
rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

  • powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
  • cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
  • codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
  • norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
  • fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
  • Mass of solvent = density x volume, i.e., \boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg  \ }

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0
3 years ago
Read 2 more answers
Net ionic of ammonium sulfide added to iron (ll) chloride
meriva

Answer:

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)  

Step-by-step explanation:

Molecular Equation:

(NH₄)₂S(aq) + FeCl₂(aq) ⟶ 2NH₄Cl(aq) + FeS(s)

Ionic equation :

2NH₄⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2Cl⁻(aq) ⟶ 2NH₄⁺(aq) + 2Cl⁻(aq) + FeS(s)

Net ionic equation :

Cancel all ions that appear on both sides of the reaction arrow (underlined).

<u>2NH₄⁺(aq)</u> + S²⁻(aq) + Fe²⁺(aq) + <u>2Cl⁻(aq)</u> ⟶ <u>2NH₄⁺(aq) </u>+ 2<u>Cl⁻(aq) </u>+ FeS(s)

Fe²⁺(aq) + S²⁻(aq )⟶ FeS(s)

4 0
3 years ago
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