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alexgriva [62]
3 years ago
9

Object A has a molar heat of 31.2 J/mole∙°C and object B molar heat is 11.2 J/mole∙°C. Which object will heat up faster if they

have the same mass and equal amount of heat is applied? Explain why.
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

Substance B

Explanation:

Molar heat of A = 31.2J/mole.°C

Molar heat of B =  11.2 J/mole∙°C.

The molar heat of a substance is the amount of heat that must be added to a mole of a substance to raise the temperature by 1°C.

  • Substance B will heat up faster compared to A.
  • It has a smaller molar heat compared to A.
  • This suggests that it will require lesser heat to raise its temperature by  1°C.
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A gas with a volume of 4.0L at a pressure of 205 kPa is allowed to expand to a volume of 12000 mL. What is the pressure in atmos
Leni [432]

Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 205 kPa

P_2 = final pressure of gas = ?

V_1 = initial volume of gas  = 4.0 L

V_2 = final volume of gas = 12000 ml = 12 L    (1L=1000ml)  

205\times 4.0=P_2\times 12  

P_2=68.3kPa=0.674atm        (1kPa=0.0098atm)

Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.

8 0
3 years ago
Sodium hydroxide (NaOH) and elemental sodium (Na)
Olegator [25]

Answer:

If the question is which can make a buffer, then NH3, NH4Cl should be correct. Because Ammonium (NH4) is conjugate acid of NH3 so they can form an equilibrium which is basically a buffer whose purpose is to resist pH change.

Explanation:

6 0
2 years ago
Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.
BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>

Now we can find percentage composition / percentage by mass of oxygen.

% composition = \frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound} × 100

% composition = \frac{144}{240.9} × 100 = <u>59.776%</u>

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0
2 years ago
Consider the structures of vinegar and triglyceride and draw in any bond dipoles that exist in the molecules above using a dipol
bagirrra123 [75]

Answer:

Given molecules are vinegar and triglycerides.

Explanation:

The dipole is a vector quantity and it is heading from less electronegative atom to more electronegative atom in a polar covalent bond.

The structures and the bond dipoles in the given molecules are shown below:

5 0
3 years ago
A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25
mylen [45]
<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

  • Concentration of stock solution as 1.00 M
  • Volume of the stock solution as 50 mL
  • Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

  • The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
  • Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

     = (1.00 M × 50 mL) ÷ 0.25 M

     = 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

5 0
3 years ago
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