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3 years ago

If 7.35 L of gas at 17°C and 3.55 atm is compressed at a pressure of 123 atm and 34°C, calculate the

Chemistry

1 answer:

MaRussiya [10]3 years ago

6 0

Answer: V2= 0.224L

Explanation:

P1=3.55atm, V2= 7.35L, T1= 290, P2= 123atm, V2=?, T2= 307K

Applying general gas equation

P1V1/T1= P2V2/T2

Substitute and Simplify

3.55×7.35/290 = 123×V2/307

V2= 0.224L

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Pl help it’s for a grade I will give you Brainly

Inessa [10]

Answer:

D. B, C, A

Explanation:

A. For object A, given the following data;

Mass = 2kg

Net force = 1N

To find the acceleration;

Acceleration = net force/mass

Acceleration = 1/2

Acceleration = 0.5m/s²

B. For object B, given the following data;

Mass = 8kg

Net force = 10N

To find the acceleration;

Acceleration = net force/mass

Acceleration = 10/8

Acceleration = 1.25m/s²

C. For object C, given the following data;

Mass = 7kg

Net force = 7N

To find the acceleration;

Acceleration = net force/mass

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Therefore, placing the objects in decreasing order, according to the magnitude of the acceleration they are experiencing is B, C, A

8 0

3 years ago

Need the helps thoooo

Maslowich

Hey dude. It's letter D. Compound. Compounds are composed of two or more separate elements.

6 0

3 years ago

Read 2 more answers

Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5

77julia77 [94]

Answer: The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

Explanation:

We are given:

Concentration of buffer = 0.460 M

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$ .....(1)

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COOH]=0.460M$

$[C_6H_5COO^-]=0.460M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2$

When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.

pH change for loosing buffer capacity = [4.2 - 1] = 3.2

Calculating the ratio of conjugate base and its acid by using equation 1:

$pK_a$ = 4.2

pH = 3.2

Putting values in equation 1, we get:

$3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(2)

For benzoic acid and its conjugate base:

Molarity of benzoic acid and its conjugate base = 0.460 M

Volume of solution = 115 mL

Putting values in equation 2, we get:

$0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Let the moles of acid added to carry out the change is 'x' moles

Calculating the moles of acid added:

$\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434$

Calculating the volume of acid added by using equation 2, we get:

Moles of acid added = 0.0434 moles

Molarity of solution = 0.390 M

Putting values in equation 2, we get:

$0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL$

Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

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3 years ago

Fifteen points! Nothing like chemistry quizzes in the morning... >o<

sertanlavr [38]

Answer:

Explanation:

HA(aq)+H2O(l)⟺H3O+(aq)+A−(aq)(1)

you need to solve for the Ka value. To do that you use

Ka=[H3O+][A−][HA](2)

Another necessary value is the pKa value, and that is obtained through pKa=−logKa

The procedure is very similar for weak bases. The general equation of a weak base is

BOH⟺B++OH−(3)

Solving for the Kbvalue is the same as the Ka value. You use the formula

Kb=[B+][OH−][BOH](4)

The pKb value is found through pKb=−logKb

The Kw value is found withKw=[H3O+][OH−].

Kw=1.0×10−14(5)

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3 years ago

Acts as the digestive system inside a cell. It helps to break down old or unneeded parts of the cell, and substances that have b

DIA [1.3K]

Answer:

I think it is enzymes

Explanation:

5 0

3 years ago