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3 years ago

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A piece of lead with a mass of 14.9 grams at a temperature of 92.5 degrees Celsius is dropped into an insulated container of wat

er. The mass of the water is 165 g and its temperature before adding the lead is 20.00 degrees Celsius. After adding the lead to the water, the water temperature rises to 20.20 degrees Celsius. What is the specific heat of the lead?

2 answers:

Vesna [10]3 years ago

6 0

the specific heat of lead is

soldi70 [24.7K]3 years ago

6 0

Answer:

the answer is .128 celsius i hope this help

if it did can u mark brainliest

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Help me please I need to pass.

nlexa [21]

Answer:

2 C Atoms

Explanation:

When you have coefficient of 2 next to a compound element, it indicates there are 2 of each compound element. In the compound element, there is one C Atom, and 2×1 is 2.

4 0

3 years ago

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

Tom [10]

Answer:

Approximately $2.53\; \rm L$ (rounded to three significant figures) assuming that ${\rm HCl}\, (aq)$ is in excess.

Explanation:

When ${\rm HCl} \, (aq)$ and ${\rm AgNO_3}\, (aq)$ precipitate, ${\rm AgCl} \, (s)$ (the said precipitate) and $\rm HNO_3\, (aq)$ are produced:

${\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)$ (verify that this equation is indeed balanced.)

Look up the relative atomic mass of $\rm Ag$ and $\rm Cl$ on a modern periodic table:

$\rm Ag$ : $107.868$ .
$\rm Cl$ : $35.45$ .

Calculate the formula mass of the precipitate, $\rm AgCl$ :

$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

Notice that in the balanced equation for this reaction, the coefficients of ${\rm AgNO_3} \, (aq)$ and ${\rm AgCl}\, (s)$ are both one.

In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

4 0

3 years ago

How many moles of CO2 are produced from 16 grams of O2

Liula [17]

Answer:

One mole of CO2 has mass of 44 g and 32 g of O2. So 16 g of O2 have 22 g of CO2 or 0.5 moles of it.

brainliest plz

Explanation:

8 0

3 years ago

If a chemist wants to make 1.3 L of 0.25 M solution of KOH by diluting a stock solution of 0.675 M KOH, how many milliliters of

anzhelika [568]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

.675 M x V1 = .25 M x 1.3 L

V1 = 0.48 L or 480 mL

8 0

3 years ago

How many moles of water 9.31 x 10^22

san4es73 [151]

No calculator will show the answer

5 0

3 years ago