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lilavasa [31]
2 years ago
7

1 What direction do molecules move in?

Chemistry
1 answer:
notsponge [240]2 years ago
3 0

Answer:

The majority of the molecules move from higher to lower concentration, although there will be some that move from low to high. The overall (or net) movement is thus from high to low concentration.

hope this helps!<3

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
2 years ago
Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the
kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH  = 3.40+ log {0.170 /0.105}

pH =  3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol  = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

4 0
2 years ago
Read 2 more answers
23. What are the correct formulas and coefficients for the products of the following double replacement reaction? Al(OH)3 + H3PO
NikAS [45]
23. The correct answer would be a (AlPO4 + H2O).

This is an example of a neutralization reaction (an acid and a base react to form water and a salt).

24. a, combination reaction. There are two reactants and one product.
3 0
3 years ago
An experiment requires 0.52 M NH3(aq). The stockroom manager estimates that 15 L of the base is needed. What volume of 15 M NH3(
Bingel [31]

Answer: 0.52 L of 15 M NH_3(aq) will be used to prepare this amount of 0.52 M base.

Explanation:

But on diluting the number of moles remain same and thus we can use molarity equation.

C_1V_1(stock)=C_2V_2 (to be prepared)

where,

C_1 =concentration of stock solution = 15 M

V_1 = volume of stock solution = ?

C_2 = concentration of solution to be prepared = 0.52 M

V_2 = volume of solution to be prepared = 15 L

15\times V_1=0.52\times 15

V_1=0.52L

Thus 0.52 L of 15 M NH_3(aq) will be used to prepare this amount of 0.52 M base

8 0
2 years ago
N² +H² -&gt; NH³<br> When the equation is balanced, what is the coefficient for hydrogen gas?
Andrej [43]

Answer:

I don't even know I'm sorry

6 0
2 years ago
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