All categories

3 years ago

How much heat would be released by 1.3 kg of water if it's temperature went from 295 k to 274 k

Chemistry

1 answer:

kicyunya [14]3 years ago

4 0

Answer: The energy released will be 114.223 kJ.

Explanation:

To calculate the amount of heat released, we use the formula:

$Q= m\times c\times \Delta T$

Q = heat gained or released

m = mass of the substance = 1.3kg = 1300 g (Conversion factor: 1kg = 1000g)

c = heat capacity of water = 4.184 J/g K

$\Delta T={\text{Change in temperature}}=(274-295)K=-21K$

Putting values in above equation, we get:

$Q=1300g\times 4.184J/gK\times (-21)K$

Q = -114223 Joules = -114.223 kJ (Conversion Factor: 1kJ = 1000J)

Negative sign implies heat released.

Hence, heat released by 1.3 kg of water is 114.223kJ.

You might be interested in

Carbon naturally occurs in two forms: diamond and graphite. Why do these two forms have very different properties?

Nookie1986 [14]

The differences in the properties of diamond and graphite is as a result of how their particles are arranged in space. This space arrangement leads to distinct crystal structures for the two compounds. In diamond, the carbon atoms are arranged in tetrahedral shape while in graphite the carbon atoms arrayed in planes.
Hope this helps :)

5 0

3 years ago

How many cubic miles are 8.48E+08 gallons of water? The density of water at ambient conditions is 1.000 g/mL.

Inessa05 [86]

The volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .

<h3>Weight of one gallon of water</h3>

The weight of 1 gal of water is given as 3785 g

Mass of 8.48 x 10⁸ gal = 3785 x 8.48 x 10⁸ = 3.2 x 10¹² g

<h3>Volume of the water in cubic meters</h3>

Volume = mass/density

Volume = 3.2 x 10¹² g/1 gmL

Volume = 3.2 x 10¹² mL x 10⁻⁶ m³/mL = 3.2 x 10⁶ m³

Thus, the volume of the water in cubic meter is determined as 3.2 x 10⁶ m³ .

Learn more about volume here: brainly.com/question/1972490

#SPJ1

7 0

2 years ago

What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and N

Alecsey [184]

Answer:

Cr(OH)2(s), Na+(aq), and NO3−(aq)

Explanation:

Let is consider the molecular equation;

2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)

This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.

Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.

4 0

4 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago

A particle that orbits the nucleus in an atom is called a(n

Morgarella [4.7K]

Particles that orbit the nucleus are called electrons.

Explanation: Electrons are negatively charged particles arranged in orbits around the nucleus of an atom and determining all of the atom's physical and chemical properties except mass and radioactivity.

7 0

3 years ago

Read 2 more answers