How does the creation of oil demonstrate conservation of mass?

27. How much energy is required to change 150.0 g of water from 10.0°C to 45.0°C? (Cwater =

Alekssandra [3086]

Answer:

The correct option is C. 21900.3. I calculated 21945 J, which makes option C closely aligned with my result.

Explanation:

Data

mass = 150 g

initial temperature T1 = 10°C

final temperature T2 = 45°C

Cw = 4.18 J/g°C

Formula

Q = mCΔT = mC(T2 - T1)

Substitution

Q = (150)(4.18)(45 - 10)

Simplification

Q = (150)(4.18)(35)

Result

Q = 21945 J

5 0

2 months ago

"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

lorasvet [2795]

Answer: The percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

From the information provided:

Let the fractional abundance for $_{14}^{28}\textrm{Si}$ isotope be 'x'

For $_{14}^{28}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 27.9769 amu

Percentage abundance of $_{14}^{28}\textrm{Si}$ isotope = 92.22 %

Fractional abundance for $_{14}^{28}\textrm{Si}$ isotope = 0.9222

For $_{14}^{29}\textrm{Si}$ isotope:

Mass of $_{14}^{28}\textrm{Si}$ isotope = 28.9764 amu

Percentage abundance for $_{14}^{28}\textrm{Si}$ isotope = 4.68%

Fractional abundance of $_{14}^{28}\textrm{Si}$ isotope = 0.0468

For $_{14}^{30}\textrm{Si}$ isotope:

Mass of $_{14}^{30}\textrm{Si}$ isotope = 29.9737 amu

Fractional abundance for $_{14}^{30}\textrm{Si}$ isotope = x

The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

$28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309$

To convert this fractional abundance into a percentage, multiply by 100:

$\Rightarrow 0.0309\times 100=3.09\%$

This shows that the percentage abundance for $_{14}^{30}\textrm{Si}$ isotope is 3.09 %.

6 0

3 months ago

At –45oC, 71 g of fluorine gas take up 6843 mL of space. What is the pressure of the gas, in kPa?

KiRa [2933]

The quantity of fluorine in moles is calculated as 71/19 = 3.74
We also know that at standard temperature and pressure (273 K and 101.3 kPa), one mole of gas occupies 22.4 liters
So, the volume for 3.74 moles at S.T.P is: 3.74 x 22.4
This results in a volume of 83.776 L, which is equivalent to 83,776 mL

Next, applying Boyle's law, which states that for a fixed amount of gas,
PV = constant

We set up the equation P x 6843 = 101.3 x 83776
Solving for P gives us 1,240 kPa

4 0

3 months ago

How many kilojoules of heat are required to heat 1.37kg of water from 21.3 to 89.5

VMariaS [2998]

Q=? kJ
m= 1.37 kg × $\frac{1000 g}{1kg}$ = 1370 g

c= 4.18 J/g° C
ΔT= <span>89.5-21.3= 68.2˚C
</span>? J = (1370 g) (4.18 J/g˚C) <span>(68.2˚C)
</span>q = 390,554 J × $\frac{1kJ}{1000J}$

q = 391 kJ

5 0

3 months ago