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3 years ago

How much energy is needed to raise the temperature of a 200 g piece of lead by 250 o C?

Chemistry

1 answer:

devlian [24]3 years ago

4 0

<u>Answer:</u> The amount of energy required to raise the given amount of temperature is 6400 J

<u>Explanation:</u>

To calculate the heat absorbed by the lead, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of lead = 200 g

c = heat capacity of lead = 0.128 J/g°C

$\Delta T$ = change in temperature = 250°C

Putting values in above equation, we get:

$q=200g\times 0.128J/g^oC\times 250^oC=6400J$

Hence, the amount of energy required to raise the given amount of temperature is 6400 J

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Calculate the volume of 4.00 molar NaOH solution required to prepare 100 mL of a 0.500 molar solution of NaOH.

Kazeer [188]

Answer:

The answer to your question is V₁ = 12.5 ml

Explanation:

Data

Volume = V₁?

[NaOH] = C₁ = 4.0 M

Volume 2 = V₂ = 100 ml

[NaOH] = C₂ = 0.5 M

Formula of dilution

V₁C₁ = V₂C₂

Solve for V₁ (original solution)

V₁ = $\frac{V2C2}{C1}$

Substitution

V₁ = $\frac{(0.5)(100)}{4}$

Simplification

V₁ = $\frac{50}{4}$

Result

V₁ = 12.5 ml

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3 years ago

The number of dots around an element symbol in a Lewis dot structure equals the total number of electrons in an atom of that ele

strojnjashka [21]

The number of dots represents the amount of valence electrons, which is the same as the last digit of the elements group number in ptof.

8 0

3 years ago

A container of gas has a volume of 3.5 L and a pressure of 0.8 atm. Assuming the temperature remains constant, what volume of ga

strojnjashka [21]

Answer:

The correct answer is option A.

Explanation:

Initial volume of the gas = $V_1=3.5 L$

Final volume of the gas = $V_2=?$

Initial pressure of the gas = $P_1=0.8 atm$

Final volume of the gas = $P_2=0.5 atm$

Using Boyle's law:

$P_1V_1=P_2V_2$

$0.8 atm \times 3.5 L=0.5 atm\times V_2$

$V_2=5.6 L$

Hence,the correct answer is option A.

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3 years ago

Read 2 more answers

How many grams of NaF form when .5 mol of HF reacts with excess Na2SiO3?

IgorLugansk [536]

Answer:

5.25g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

Na2SiO3 + 8HF → H2SiF6 + 2NaF + 3H2O

From the balanced equation above,

8 moles of HF reacted to produce 2 moles of NaF.

Therefore, 0.5 moles of HF will react to produce = (0.5 x 2)/8 = 0.125 mole of NaF.

Next, we shall convert 0.125 mole of NaF to grams.

This is illustrated below:

Mole of NaF = 0.125 mole

Molar mass of NaF = 23 + 19 = 42g/mol

Mass of NaF =..?

Mass = mole x molar mass

Mass of NaF = 0.125 x 42

Mass of NaF = 5.25g

Therefore, 5.25g of NaF is produced from the reaction.

7 0

3 years ago

calculate how many milliliters of 0.142 M NaOH are needed to completely neutralize 21.4 mL of 0.294 M H2C4H4O6.

Flura [38]

Answer:

The answer to your question is 88.7 ml

Explanation:

Data

Volume = ?

Concentration of NaOH = 0.142 M

Volume of H₂C₄H₄O₆ = 21.4 ml

Concentration of H₂C₄H₄O₆ = 0.294 M

Balanced chemical reaction

2 NaOH + H₂C₄H₄O₆ ⇒ Na₂C₄H₄O₆ + 2H₂O

1.- Calculate the moles of H₂C₄H₄O₆

Molarity = moles/volume

Solve for moles

moles = Molarity x volume

Substitution

moles = 0.294 x 21.4/1000

Result

moles = 0.0063

2.- Use proportions to calculate the moles of NaOH

2 moles of NaOH ------------------ 1 moles of H₂C₄H₄O₆

x ------------------ 0.0063 moles

x = (0.0063 x 2) / 1

x = 0.0126 moles of NaOH

3.- Calculate the volume of NaOH

Molarity = moles / volume

Solve for volume

Volume = moles/Molarity

Substitution

Volume = 0.0126/0.142

Result

Volume = 0.088 L or 88.7 ml

3 0

3 years ago