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Rzqust [24]
2 years ago
15

A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead

is 0.129 J/g°C.
Chemistry
1 answer:
Ludmilka [50]2 years ago
3 0

Answer:

The energy required is 3225 Joules.

Explanation:

Given,

mass of lead cube = 500 grams

T₁ = 25°C

T₂ = 75°C

specific heat of lead = 0.129 J/g°C

Energy required to heat the lead can be found by using the formula,

Q = (mass) (ΔT) (Cp)

Here, ΔT = T₂ - T₁ = 75 - 25 = 50

Substituting the values,

Q = (500)(50)(0.129)

Q = 3225 Joules.

Therefore, energy required is 3225 J.

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175 mL of Cl2 gas is held in a flexible vessel at
Gnoma [55]

Answer:

V₂ = 15.6 L

Explanation:

Given data:

Initial volume = 175 mL  (0.175 L)

Initial pressure = 1 atm

Initial temperature = 273 K

Final temperature = -5°C (-5+273 = 268 K)

Final volume = ?

Final pressure = 1.16 kpa (1.16/101=0.011 atm)

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 1 atm × 0.175 L × 268 K / 273 K × 0.011 atm

V₂ = 46.9 L / 3.003

V₂ = 15.6 L

7 0
2 years ago
Read 2 more answers
To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple _______. Then, plot ln(K
Aloiza [94]

Answer:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us use the thermodynamic definition of the Gibbs free energy and its relationship with Ksp as follows:

\Delta G=-RTln(Ksp)\\\\\Delta G=\Delta H-T\Delta S

Thus, by combining them, we obtain:

-RTln(Ksp)=\Delta H-T\Delta S\\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{T\Delta S}{RT} \\\\ln(Ksp)=-\frac{\Delta H}{RT} +\frac{\Delta S}{R}

Which is related to the general line equation:

y=mx+b

Whereas:

y=ln(Ksp)\\\\m=-\frac{\Delta H}{R} \\\\x=\frac{1}{T} \\\\b=\frac{\Delta S}{R}

It means that we answer to the blanks as follows:

To determine the enthalpy and entropy of dissolving a compound, you need to measure the Ksp at multiple temperatures. Then, plot ln(Ksp) vs. 1/T. The slope of the plotted line relates to the enthalpy (ΔH) of dissolving and the intercept of the plotted line relates to the entropy (ΔS) of dissolving.

Regards!

8 0
1 year ago
Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit
Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

0.0811 M=\frac{n}{0.0017 L}

n = 0.0001378 mol

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0
2 years ago
You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl
salantis [7]

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}     .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M

Hence, the new concentration of lemonade is 3.90 M

7 0
2 years ago
At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper
fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0
2 years ago
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