So i need answers. Take your time its alright im not going to push it out. Thanks!

What causes the heat we feel on a hot, sunny day?

Nataly_w [17]

What cause the heat we feel on a hot, sunny day is UV Rays my mans

5 0

3 years ago

The word science is derived from latin word..........which means............

stepan [7]

Scientia, meaning knowledge and skill

8 0

3 years ago

Which compound has the same empirical and molecular formula ethyne ethene ethane methane?

Margaret [11]

Empirical formula is the simplest ratio of whole numbers of components in a compound
molecular formula is the actual ratio of components in a compound .
the molecular formula for the compounds given are as follows
ethyne - C₂H₂
ethene - C₂H₄
ethane - C₂H₆
methane - CH₄
the actual ratios of the elements simplified ratio
C : H C : H
ethyne 2:2 1:1
ethene 2:4 1:2
ethane 2:6 1:3
methane 1:4 1:4
the only compound where the actual ratio is equal to the simplified ratio is methane
therefore in methane molecular formula CH₄ is the same as empirical formula CH₄

6 0

3 years ago

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

3 years ago

Never mind jhvjycdtrsesetdfyguhbjnk

OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial pressure of methanol is $P_{CH_3OH_{(g)}} =0.077 \ bar$

The partial pressure of carbon monoxide is $P_{CO} = 0.382 \ bar$

The partial pressure at hydrogen is $P_H = O \ bar$

Explanation:

From the question we are told that

The volume of the flask is $V_f = 10.0 \ L$

The initial pressure of carbon monoxide gas is $P_{CO} = 0.461 \ bar$

The initial temperature of carbon monoxide gas is $T_{CO} = 22.0^oC$

The volume of the hydrogen gas is $V_h = 200 mL = 200 *10^{-3} \ L$

The initial pressure of the hydrogen is $P_H = 7.10 \ bar$

The initial temperature of the hydrogen is $T_H = 271 \ K$

The reaction of carbon monoxide and hydrogen is represented as

$CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}$

Generally from the ideal gas equation the initial number of moles of carbon monoxide is

$n_1 = \frac{P_{CO} * V_f }{RT_{CO}}$

Here R is the gas constant with value $R = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K$

=> $n_1 = \frac{0.461 * 10 }{0.0821 * (22 + 273)}$

=> $n_1 = 0.19$

Generally from the ideal gas equation the initial number of moles of Hydrogen is

$n_2 = \frac{P_{H} * V_H }{RT_{H}}$

$n_2 = \frac{ 7.10 * 0.2 }{0.0821 * 271 }$

=> $n_2 = 0.064$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of CO

=> 0.064 moles of hydrogen gas will react with x mole of CO

So

$x = \frac{0.064}{2}$

=> $x = 0.032 \ moles \ of \ CO$

Generally from the chemical equation of the reaction we see that

2 moles of hydrogen gas reacts with 1 mole of $CH_3OH_{(g)}$

=> 0.064 moles of hydrogen gas will react with z mole of $CH_3OH_{(g)}$

So

$z = \frac{0.064}{2}$

=> $z = 0.032 \ moles \ of \ CH_3OH_{(g)}$

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is

$n_k = n_1 - x$

=> $n_k = 0.19 - 0.032$

=> $n_k = 0.156$

So at the end of the reaction , the partial pressure for CO is mathematically represented as

$P_{CO} = \frac{n_k * R * T_{CO}}{V}$

=> $P_{CO} = \frac{0.158 * 0.0821 * 295}{10}$

=> $P_{CO} = 0.382 \ bar$

Generally the partial pressure of hydrogen is 0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial pressure of the methanol is mathematically represented as

$P_{CH_3OH_{(g)}} = \frac{z * R * T_{CO}}{V_f}$

Here $T_{CO}$ is used because it is given the question that the temperature returned to 22.0 degrees C

So

$P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 * 295}{10}$

$P_{CH_3OH_{(g)}} =0.077 \ bar$

6 0

3 years ago

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