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3 years ago

How does the density of a hershey bar compare to the density of the a half bar?

1 answer:

8 0

The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.

For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml

Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.

See, although there are different amounts of the same substance, their density is the same.

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The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles$

b) moles of $C_2H_4$

$\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ combine with 1 mole of $H_2$

Thus 0.33 mole of $C_2H_4$ will combine with = $\frac{1}{1}\times 0.33=0.33$ mole of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product.

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.33 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.33=0.33moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g$

Thus theoretical yield (g) of $C_2H_6$ produced by the reaction is 9.9 grams

7 0

3 years ago

What is the balanced chemical equation for this reaction? H3PO4 + HCl → PCl5 + H2O Question 7 options: H3PO4 + 5HCl → PCl5 + H2O

WITCHER [35]

Answer:

It's the last option.

Explanation:

H3PO4 + 5HCl → PCl5 + 4H2O

4 0

3 years ago

Technetium-104 has a half life of 18.0 minutes. how much of a 165.0 g sample remains after 90.0 minutes?

11Alexandr11 [23.1K]

The mass of 165.0 g sample that remains after 90.0 minutes is 5.16 grams

calculation

lambda㏑2/18= 0.0385

m(t)= 165 x e( 0.0385 x90) =5.16g

8 0

3 years ago

Read 2 more answers

What law is known as the law of action- reaction?

Molodets [167]

Law of Action and Reaction mean "To every action there is equal and opposite reaction.

Explanation:

It is Newtons third law which explained few things such as :-

1. The nature of the forces between the two interacting objects.

2. According to this law , the force that is exerted from Object-1 to Object-2 is said to be equal in magnitude and opposite to the direction to the force exerted to object 2 upon object 1.

3. Eaxmple - As we push against the wall - Action , the wall in return opposes equal force to resist the pressure of the hand - Reaction.

6 0

3 years ago

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the

fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

b) Br2 is the limiting reactant.

c) There will be formed 86.3 grams of BrF3

d) There will remain 0.326 moles of F2 = 12.97 grams

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

Br2 is the limiting reactant. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = 0.326 moles of F2 = 12.97 grams

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = 86.3 grams

7 0

2 years ago