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2 years ago

If 0.959J of heat is added to 0.88g of water, how much will its temperature increase

Chemistry

1 answer:

ss7ja [257]2 years ago

7 0

Answer:

0.260 Celsius

Explanation:

q =c x m x (T2-T1)

c - specific heat of water 4.186 J/g.C

T2-T1 = q /(c x m) = 0.959 /(4.186 x 0.88) = 0.959/3.68 =0.260 C

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Use Boyle's Law to solve the following problem. A 400 mL sample of a gas is at 10

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Answer:

v2 = 100 Ml OR 0.1 Liters

Explanation:

(400 Ml) (10 atm) = (v2) (40 atm)

4000 = 40v2

100 Ml = v2

Liters = ml / 1000

Liters = 100 / 1000

Liters = 0.1

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3 years ago

Examine the analogy and how it compares a part to a whole. Link : chain :: atom : _[blank]_ what option best completes this ana

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3 years ago

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Quinine is an important type of molecule that is involved in photosynthesis. The transport of electrons mediated by quinone in c

Irina18 [472]

Answer:

The heat of combustion is -25 kJ/g = -2700 kJ/mol.

Explanation:

According to the Law of conservation of energy, the sum of the heat released by the combustion reaction and the heat absorbed by the bomb calorimeter is equal to zero.

Qcomb + Qcal = 0

Qcomb = - Qcal

The heat absorbed by the calorimeter can be calculated with the following expression.

Qcal = C × ΔT

where,

C is the heat capacity of the calorimeter

ΔT is the change in temperature

Then,

Qcomb = - Qcal

Qcomb = - C × ΔT

Qcomb = - 1.56 kJ/°C × 3.2°C = -5.0 kJ

Since this is the heat released when 0.1964 g o quinone burns, the energy of combustion per gram is:

$\frac{-5.0kJ}{0.1964g} =-25kJ/g$

The molar mass of quinone (C₆H₄O₂) is 108 g/mol. Then, the energy of combustion per mole is:

$\frac{-25kJ}{g} .\frac{108g}{1mol} =-2700kJ/mol$

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3 years ago

One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c

Alchen [17]

Answer:

Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of $\rm O_2\; (g)$ is larger than that of $\rm H_2\; (g)$ (by a factor of about $16$ .) Therefore, the mass of the $\rm O_2\; (g)$ sample is significantly larger than that of the $\rm H_2\; (g)$ sample.

Explanation:

The $\rm O_2\; (g)$ and the $\rm H_2\; (g)$ sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles ( $\rm O_2\; (g)$ and $\rm H_2\; (g)$ molecules, respectively.) That is:

$n(\mathrm{O_2}) = n(\mathrm{H}_2)$ .

Note that the mass of a gas $m$ is different from the number of gas particles $n$ in it. In particular, if all particles in this gas have a molar mass of $M$ , then:

$m = n \cdot M$ .

In other words,

$m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2})$ .
$m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2})$ .

The ratio between the mass of the $\rm O_2\; (g)$ and that of the $\rm H_2\; (g)$ sample would be:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}$ .

Since $n(\mathrm{O_2}) = n(\mathrm{H}_2)$ by Avogadro's Law:

$\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ .

Look up relative atomic mass data on a modern periodic table:

$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Therefore:

$M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}$ .
$M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}$ .

Verify whether $\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}$ :

Left-hand side: $\displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1$ .
Right-hand side: $\displaystyle \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}= \frac{31.998\; \rm g \cdot mol^{-1}}{2.016\; \rm g \cdot mol^{-1}} \approx 15.9$ .

Note that the mass of the $\rm H_2\; (g)$ sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

7 0

3 years ago

An aqueous solution of a certain chemical is found to boil at 104.5 oC. At what temperature is this solution likely to freeze? T

xeze [42]

Answer:

16.4 °C

Explanation:

Boiling point elevation is the phenomenon in which the boiling point of a solvent will increase when another compound is added to it; meaning that athe resultant solution has a higher boiling point than its pure solvent.

Using the ebullioscopic constant,

ΔT = m * i * Kb

Where,

Δ T is the temperature difference between the boiling point of the solution, Temp.f and boiling point of the pure solvent, Temp.i

Kb is the ebulliscope factor of water = 0.510 °C.kg/mol

i is the van hoffs number = 1

m is the molality in mol/kg.

Calculating the molality of the solution,

Temp.i = 100°C

Temp.f = 104.5 °C

= 4.5/(1*0.510)

= 8.8235 mol/kg

Freezing point depression is defined as the decrease in the freezing point of a solvent on the addition of a solute.

Using the same equation, but kf = 1.86 °C.kg/mol

ΔT = m * i * Kf

Temp.i = freezing point of water = 0°C

Temp.f = (8.8235*1.86) - 0

= 16.412 °C

Freezing point of the solution = 16.4 °C

3 0

2 years ago