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GenaCL600 [577]
3 years ago
8

At the candy store, gourmet jelly beans sell for $1.15 per ounce. There are 28 students in Alfred's class. If each student eats

0.035 kg of jelly beans, how much would it cost to feed the class a gourmet jelly bean treat?
Chemistry
1 answer:
saw5 [17]3 years ago
8 0

Answer:

Total cost = $40.25 (Approx)

Explanation:

Given:

Per ounce = $1.15

Number of student = 28

Each student eat = 0.035 kg

Find:

Total cost

Computation:

Total weight of candy = 28 × 0.035 kg

Total weight of candy = 0.98 kg

1 ounce = 0.028 kg (approx).

Total weight of candy = 0.98 kg / 0.028

Total weight of candy = 35 ounce (Approx)

Total cost = 35 × $1.15

Total cost = $40.25 (Approx)

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A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse
ludmilkaskok [199]

Answer:

\large \boxed{\text{5.9 s}}

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases, the ratio of their rates of effusion is

\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

The time for diffusion is inversely proportional to the rate.

\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}

4 0
3 years ago
A sample of nitrogen gas collected at a pressure of 1.03 atm and a temperature of 279 K is found to occupy a volume of 568 milli
DIA [1.3K]

Answer: 0.025 moles of nitrogen gas are there in the sample.

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 1.03 atm

V = Volume of gas = 568 ml = 0.568 L   (1L=1000ml)

n = number of moles  = ?

R = gas constant =0.0821Latm/Kmol

T =temperature =279K

n=\frac{PV}{RT}

n=\frac{1.03atm\times 0.568L}{0.0821L atm/K mol\times 279K}=0.025moles

0.025 moles of nitrogen gas are there in the sample.

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What would happen if a roof was made from wool?
vichka [17]

Answer:

B

Explanation:

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Will give free brainlist !! :)
luda_lava [24]

Answer:

C is the correct answer

Explanation:

7 0
3 years ago
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