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3 years ago

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Debscribe each of the following with the correct term

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Irina-Kira [14]3 years ago

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Which is a symbol that represents SI units for temperature.

dsp73

Answer:

A.'C

Explanation:

Please answer my question

5 0

4 years ago

Read 2 more answers

A 75.0 g sample of dinitrogen monoxide is confined in a 3.q L vessel. What is the pressure( in atm) at 115 celsius

Nonamiya [84]

Data Given:
Pressure = P = ?

Volume = V = 3.0 L

Temperature = T = 115 °C + 273 = 388 K

Mass = m = 75.0 g

M.mass = M = 44 g/mol

Solution:
Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ------ (1)
Calculating Moles,
n = m / M

n = 75.0 g / 44 g.mol⁻¹

n = 1.704 mol

Putting Values in Eq. 1,

P = (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

P = 18.08 atm

7 0

3 years ago

over a 12.3 minuete period 5.13 E-3 moles of F2 gas effuses from a contaier. How many moles of CH4 gas could effuse from from th

Aleonysh [2.5K]

Answer : The moles of methane gas could be, $7.90\times 10^{-3}mol$

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$

$[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}$

where,

$R_1$ = rate of effusion of fluorine gas

$R_2$ = rate of effusion of methane gas

$n_1$ = moles of fluorine gas = $5.13\times 10^{-3}mol$

$n_2$ = moles of methane gas = ?

$t_1=t_2$ = time = 12.3 min (as per question)

$M_1$ = molar mass of fluorine gas = 38 g/mole

$M_2$ = molar mass of methane gas = 16 g/mole

Now put all the given values in the above formula 1, we get:

$[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}$

$n_2=7.90\times 10^{-3}mol$

Therefore, the moles of methane gas could be, $7.90\times 10^{-3}mol$

8 0

3 years ago

A 11.630 g milk chocolate bar is found to contain 7.815 g of sugar.How many milligrams of sugar does the milk chocolate bar cont

zzz [600]

3.815 that's the awnser......................................

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3 years ago

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Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that

iragen [17]

Answer:

108.43 grams KNO₃

Explanation:

To solve this problem we use the formula:

ΔT = Kf * b * i

Where

ΔT is the temperature difference (14.5 K)

Kf is the cryoscopic constant (1.86 K·m⁻¹)

b is the molality of the solution (moles KNO₃ per kg of water)

and<em> i</em> is the van't Hoff factor (2 for KNO₃)

We <u>solve for b</u>:

14.5 K = 1.86 K·m⁻¹ * b * 2

b = 3.90 m

Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:

275 mL water ≅ 275 g water

275 g /1000 = 0.275 kg

moles KNO₃ = molality * kg water = 3.90 * 0.275

moles KNO₃ = 1.0725 moles KNO₃

Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:

1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃

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3 years ago