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4 years ago

#1: Which of the following statements is true?

2 answers:

6 0

<span>Endothermic reactions take in heat from the environment in order to react while exothermic reactions release heat during the reaction. For this, D must be true as the products have less energy because they have released heat during the reaction. A and C are incorrect because they products and reactants are equal and B is wrong because the energy of the products would be more as the reaction took in heat.</span>

wlad13 [49]4 years ago

4 0

Answer: Option (D) is the correct answer.

Explanation:

In an endothermic reaction energy is absorbed by reactant molecules in a chemical reaction as a result energy of products is more than the energy of reactants.

Whereas in an exothermic reaction energy is released by a chemical reaction as a result energy of reactants is more than the energy of products.

Therefore, we can conclude that the correct statement is in an exothermic reaction, the energy of the products is less than the energy of the reactants.

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Michelle is trying to find the average atomic mass of a sample of an unknown

GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

Abundance of isotope A (A%) = 59.34%
Mass of isotope A = 113.6459 amu
Mass of isotope B = 115.8488 amu
Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

$Average \: atomic \: mass \: = \frac{mass \: of \: A \times A\%}{100} + \frac{mass \: of \: B \times B\%}{100} \\ \\ Average \: atomic \: mass \: = \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\ \\ Average \: atomic \: mass \: = 114.54 \: amu \\ \\$

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0

3 years ago

What is the boiling point of 0.464 m lactose in water? (Kb of water = 0.512 oC/m). Enter your answer to 3 decimal places.

kirill [66]

Answer:

Boiling point for the solution is 100.237°C

Explanation:

We must apply colligative property of boiling point elevation

T° boiling solution - T° boiling pure solvent = Kb . m

m = molalilty (a given data)

Kb = Ebulloscopic constant (a given data)

We know that water boils at 100°C so let's replace the information in the formula.

T° boiling solution - 100°C = 0.512 °C/m . 0.464 m

T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C

3 0

3 years ago

Sulphur +_________ gives ➙ sulphur oxide

Luba_88 [7]

Answer:

Sulphur + oxygen gives ➙ Sulphur oxide

Explanation:

Sulfur oxide is a compound that consists of sulfur and oxygen molecules. I hope this helps!

3 0

3 years ago

Read 2 more answers

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0

3 years ago

If 1.16 L of water is initially at 24.2 ∘C, what will its temperature be after absorption of 9.4×10−2 kWh of heat?

vitfil [10]

Answer:

The temperature will be 93.92 °C

Explanation:

To explain this we will use following equation: also Q = ∆U + W known as the NON-FLOW ENERGY EQUATION (N.F.E.E.)

With Q = heat added to the system

with ∆U = change in internal energy

⇒∆U = ( m )( Cv )( T2 - T1 )

With W = work done by the system

⇒For this situation W = 0 because there isn't work done

So we get: ∆U = ( m )( Cv )( T2 - T1 ) = Q

To find the temperature, we have to isolate T in the equation:

(T2-T1) = Q / (m)(Cv)

⇒ Since we know that m = density * volume we can calculate the mass of water.

mass = 1000g/L * 1.16 L = 1160g

Cv = heat capacity ⇒ water has a heat capacity of 4.184 J/g °C

We know the absorption of heat is 9.4x 10^-2 kWh but to know how many joule this is we should convert ( 1 joule = 3.6 x 10^6 kWh)

⇒Q = ( 0.094 kWh ) ( 3.6 x 10^6 J / kWh ) = 0.3384 x 10^6 J

For the temperature we get then: T2 -T1 = Q / (m)(Cv)

T2 - T1 = 0.3384 x 10^6 J / (( 1160g)*(4.184 J/g °C)) = 69.72 ° C

T2 = ( T2 - T1 ) + T1 ⇒ 69.72 + 24.2 = 93.92 °C

6 0

3 years ago