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Dmitry [639]
2 years ago
8

Predict the products and then balance the equation

Chemistry
2 answers:
Aneli [31]2 years ago
4 0

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the balanced equation for given Double displacement reaction ~

\sf Pb(NO_3)_2 +2 \:  KI = PbI_2  +2 \:  KNO_3

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )

IceJOKER [234]2 years ago
4 0

\\ \rm\Rrightarrow Pb(NO_3)_2+2KI\longrightarrow  2KNO_3+PbI_2

On both sides

  • Pb=1
  • NO_3=2
  • K=2
  • I=2
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If the heat released during condensation goes only to warming the iron block, what is the final temperature (in ∘C) of the iron
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Answer:

91°C

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree

From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.

Then number of moles =given mass/ molecular mass

Molecular mass of water= 18 g/mol

Given mass= 0.95 g

( 0.95 g/18 g/mol)

= 0.053 moles

Then Heat evolved during condensation = moles of water x Latent heat of vaporization

Q= heat absorbed or released

H=enthalpy of vaporization for water

n= number of moles

Q=nΔH

Q = 0.053 moles x 44.0 kJ/mol

= 2.322 Kj

=2322J

We can now calculate Heat gained by Iron block

Q = mCΔT

m = mass of substance

c = specific heat capacity

=change in temperature

m = 75 g

c = 0.450 J/g/°C

If we substitute into the above formula we have

Q= 75 x 0.450 x ΔT

2322 = 75 x 0.450 x ΔT

ΔT = 68.8°C

Since we know the difference in temperature, we can calculate the final temperature

ΔT = T2 - T1

T1= Initial temperature = 22°C

T2= final temperature

ΔT= change in temperature

T2 = T1+ ΔT

= 68.8 + 22

= 90.8 °C

=91°C

Therefore, final temperature is 91°C

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