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2 years ago

At 19 °C, the vapor pressure of pure ethanol is 40.0 mmHg. What is the pressure in atm? atm

Chemistry

1 answer:

m_a_m_a [10]2 years ago

5 0

Answer:

0.05263158 atm

Explanation: mmHg to Atmosphere Conversion Example. Task: Convert 975 mmHg to atmospheres (show work) Formula: mmHg ÷ 760 = atm Calculations: 975 mmHg ÷ 760 = 1.28289474 atm Result: 975 mmHg is equal to 1.28289474 atm. This is an example for how i got the answer.

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NiSe2 + Ca3(PO4)2 = Ni3(PO4)2 + CaSe<br><br> balance please

Aliun [14]

3Ni + Sn3(PO4)2 → Ni3(PO4)2 + 3Sn I think. Not for sure though

8 0

2 years ago

¿Cuál es el % m/m de una disolución en que hay disueltos 22 g de soluto en 44 g de disolvente?

Archy [21]

The question is as follows: What is the% m / m of a solution in which 22 g of solute are dissolved in 44 g of solvent?

Answer: The% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

Explanation:

Given: Mass of solute = 22 g

Mass of solvent = 44 g

The percentage m/m is calculated using the following formula.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100$

Substitute the values into above formula as follows.

$Mass percentage = \frac{mass of solute}{mass of solvent} \times 100\\= \frac{22 g}{44 g} \times 100\\= 50 percent$

Thus, we can conclude that the% m/m of a solution in which 22 g of solute are dissolved in 44 g of solvent is 50%.

5 0

2 years ago

A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L

Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{350\times 15}{35}\\\\V_2=150\ mL$

So, the new volume is 150 mL.

8 0

3 years ago

The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,

Katyanochek1 [597]

Answer:

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

$n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

$n_g = 0.00215 \ mols$

Given that the number of moles of CHCl3 evaporated is less than the number of moles of CHCl3 initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of air so the pressure at equilibrium is 100 mmHg

4 0

3 years ago

How many grams of water are made from the reaction of 16.0 grams of o2? given the reaction: 2h2 o2 --> 2h2o

Jlenok [28]

Answer:

18.0 g H₂O

Explanation:

To find the mass of water (H₂O), you need to (1) convert grams O₂ to moles O₂ (via the molar mass), then (2) convert moles O₂ to moles H₂O (via mole-to-mole ratio from equation coefficients), and then (3) convert moles H₂O to grams H₂O (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value.

Molar Mass (O₂): 2(15.998 g/mol)

Molar Mass (O₂): 31.996 g/mol

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

2 H₂ + 1 O₂ -----> 2 H₂O

16.0 g O₂ 1 mole 2 moles H₂O 18.014 g
--------------- x ---------------- x --------------------- x ----------------- = 18.0 g H₂O
31.996 g 1 mole O₂ 1 mole

6 0

2 years ago