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LUCKY_DIMON [66]
3 years ago
10

What mass of water at 10.0 oC would be required to cool 50.0g of a metal having a specific heat of

Chemistry
1 answer:
Stels [109]3 years ago
7 0
<h3>Answer:</h3>

The mass of water required is 50.2 g

<h3>Explanation:</h3>
  • Quantity of heat of a substance is calculated by multiplying the mass of a substance by the specific heat capacity and the change in temperature.

That is;

Q=m×c×ΔT

In this case, water is used to cool a metal, therefore, water will gain heat while the metal will lose heat.

Heat gained is equivalent to heat lost

Heat gained by water = Heat lost by the metal

Step 1: Heat lost by the metal

Mass of the metal  = 50.0 g

Specific heat capacity of metal = 0.60 J/g°C

Change in temperature (ΔT) = 70°C

Heat lost by the metal = 50 g× 0.6 × 70

                                     = 2100 Joules

Step 2: Heat gained by water

Mass of water = x g

Specific heat capacity of water = 4.18 J/g°C

Temperature change = 10 °C

Heat gained by water = x g × 4.18 × 10

                                    = 41.8x joules

Step 3: Mass of water

Heat gained by water = heat lost by the metal

41.8x = 2100

   x = 50.239 g

      = 50.2 g (1 d.p)

The mass of water is 50.2 g

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How many mL of a stock 50% (w/v) KNO3 solution are needed to prepare 250 mL of a 20% (w/v) KNO3 solution?
Andre45 [30]

Answer:

100ml of a stock 50% KNO3 solutions are needed to prepare 250ml of a 20% KNO3 solution.

Explanation:

In the given question it is mentioned that

     S1=50%

      V2=250ml

      S2= 20%

We all know that

                     V1S1=V2S2

                     ∴V1=  V2×S2÷S1

                     ∴V1=  V2S2×1/S1

                      ∴V1= 250×20÷50

                       ∴V1= 100ml

 

6 0
3 years ago
How many atoms in 5.01 mole of lead?
leva [86]

Answer:

30.17 × 10²³ atoms

Explanation:

Given data:

Number of moles of lead = 5.01 mol

Number of atoms = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

In given question:

1 mole = 6.022 × 10²³ atoms

5.01 mol × 6.022 × 10²³ atoms / 1 mol

30.17 × 10²³ atoms

3 0
3 years ago
Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of
pantera1 [17]

Answer:

W=5743.1077\ J

Explanation:

The expression for the work done is:

W=RT \ln \left( \dfrac{P_1}{P_2} \right)

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

T is the temperature

P₁ is the initial pressure

P₂ is the final pressure

Given that:

T = 300 K

P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)

W=300\times \:8.314\ln \left(10\right)

W=2.30258\times \:2494.2

W=5743.1077\ J

5 0
3 years ago
Collected data consists of:
san4es73 [151]

The number of moles of the magnesium (mg) is 0.00067 mol.

The number of moles of hydrogen gas is 0.0008 mol.

The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.

<h3>Number of moles of the magnesium (mg)</h3>

The number of moles of the magnesium (mg) is calculated as follows;

number of moles = reacting mass / molar mass

molar mass of magnesium (mg) = 24 g/mol

number of moles = 0.016 g / 24 g/mol = 0.00067 mol.

<h3>Number of moles of hydrogen gas</h3>

PV = nRT

n = PV/RT

Apply Boyle's law to determine the change in volume.

P1V1 = P2V2

V2 = (P1V1)/P2

V2 = (101.39 x 146)/(116.54)

V2 = 127.02 mL

Now determine the number of moles using the following value of ideal constant.

R = 8.314 LkPa/mol.K

n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)

n = 0.0008

<h3>Volume of 1 mole of hydrogen gas at STP</h3>

V = nRT/P

V = (1 x 8.314 x 273) / (101.325)

V = 22.4 L

Learn more about number of moles here: brainly.com/question/13314627

#SPJ1

7 0
2 years ago
ΔH for the formation of CuCl2 from its elements is -220.1 kJ/mol. How many kJ are associated with the formation of 0.30 mole of
otez555 [7]
0.3 * - 220.1 = - 66.03kJ
3 0
3 years ago
Read 2 more answers
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