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3 years ago

How much heat is absorbed by a 56g iron skillet when its temperature rises from 15oC to 27oC?

Chemistry

1 answer:

wel3 years ago

4 0

we know that $Q=mc_{p}$ Δ $T$

$m=56g$

° $T_{1} =15$ ° $C$

$T_{2} =27$ ° $C$

$Q=56*0.450*(27-15)=302.4J$

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2 years ago

Read 2 more answers

g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol

Ronch [10]

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$M=0.0637M$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)$

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

$n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2} *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}$

Finally, the resulting molarity in 30.8 mL (0.0308 L):

$M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M$

Regards.

3 0

2 years ago