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devlian [24]
3 years ago
10

A container holds 3.41×10−3mol of carbon dioxide (CO2). After the addition of 8.41×10−4mol of carbon dioxide, the volume of the

container increases to 95.2mL, with the temperature and pressure remaining constant. What was the initial volume of the container?
Chemistry
2 answers:
mel-nik [20]3 years ago
3 0

Answer:

76.366

Explanation:

Knewton answer is 76.4

poizon [28]3 years ago
3 0

Answer:

76.4mL

Explanation:

First, calculate the final number of moles of carbon dioxide (n2) in the container.

n2 = n1+n added = 3.41×(10^-3) mol+8.41×(10^-4) mol = 4.25×(10^-3) mol

Rearrange Avogadro's law to solve for V1.

V1 =(V2×n1)/n2

Substitute in the known values for n1, V2, and n2.

V1 = (95.2mL×(3.41×10^-3 mol))/(4.25×(10^-3) mol)=76.4mL

So the initial volume is 76.4mL.

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Evaluate the following conversion. Will the answer be correct? Explain. If incorrect, how could you adjust one of the factors to
elena55 [62]

Given :

Rate = 75 m 1 s×60 s 1 min×1 h 60 m.

To Find :

Correct answer after conversion.

Solution :

We know, 1 min = 60 sec.

1 hour = 60 min = 60×60 sec = 3600 sec.

Putting value of min and hour in seconds , we get :

R=((75\times 60) + 1 )\times ( (60 + 1)\times 60 ) \times ( 3600+3600)\ s^3\\\\R=1.186\times 10^{11} \ s^3

Hence, this is the required solution.

6 0
3 years ago
A sample of a monoprotic acid (ha) weighing 0.384 g is dissolved in water and the solution is titrated with aqueous naoh. if 30.
posledela
The formula for the monoprotic acid is taken as HA, reaction with base is as follows;
HA + NaOH ---> NaA + H₂O
Stoichiometry of acid to base is 1:1
At the neutralisation point, number of HA moles = number of base moles
Number of NaOH moles reacted = 0.100M / 1000 mL /L x 30.0 mL = 0.003 mol
Therefore number of HA moles reacted = 0.003 mol
the mass of acid 0.384 g
Therefore molar mass - 0.384 g/ 0.003 mol = 128 g/mol
3 0
3 years ago
A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4
Stella [2.4K]

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

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

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

8 0
3 years ago
1. Write the balanced chemical equation for the reaction
Step2247 [10]
#1 is already balanced
8 0
3 years ago
Explain the difference between a suspension and a solution?
Verdich [7]

Answer: A suspension is a heterogenous mixture containing large particles that will settle on standing. Sand in water is an example of a suspension. A solution is a homogenous mixture of two or more substances where one substance has dissolved the other.

6 0
4 years ago
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