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emmasim [6.3K]
3 years ago
13

Describe what happened when Ag+ combined with Na2CO3. what does this indicate?

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
4 0
When Ag+ is combined with Na2CO3, the substances formed are Ag2CO3 and Na+. In this case, Ag performs single substitution over the element Na to form another set of substances. There are other types of reactions like double displacement, decomposition, etc.
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Answer:

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2. form a hypothesis

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Explanation:

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3 years ago
Which of these terms describes the number of protons in the nucleus of an atom?
Leokris [45]
Is it B................
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100POINTS!!!!<br><br> Label each reaction to either homogeneous OR heterogeneous
liraira [26]
First is homogeneous and so is the third. second is heterogeneous
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2 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
Enter your answer in the provided box.
Lina20 [59]

Answer:

5.06atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (Litres)

V2 = final volume (Litres)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 1.34 atm

P2 = ?

V1 = 5.48 L

V2 = 1.32 L

T1 = 61 °C = 61 + 273 = 334K

T2 = 31 °C = 31 + 273 = 304K

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7.34/334 = 1.32P2/304

Cross multiply

334 × 1.32P2 = 304 × 7.34

440.88P2 = 2231.36

P2 = 2231.36/440.88

P2 = 5.06

The final pressure is 5.06atm

6 0
2 years ago
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