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lina2011 [118]
3 years ago
15

If you have 23mg of water, at what temperature will it Boil?

Chemistry
1 answer:
Ivenika [448]3 years ago
8 0

Answer:

i believe it is in the 200

Explanation:

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What 3 parts of the water cycle are a result of gravity?
Kryger [21]

Answer:

Hey mate.....

Explanation:

This is ur answer......

<em>While sunlight is the energy source, the greatest force propelling the water cycle is gravity. Gravity is the force of attraction between two objects, and Earth's gravity pulls matter downward, toward its center. It pulls precipitation down from clouds and pulls water downhill. Gravity also moves air and ocean water.</em>

Hope it helps!

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7 0
3 years ago
Acids and bases:
Papessa [141]
Acids and bases ionoze the water. so it would have to be A
6 0
3 years ago
Read 2 more answers
If the initial rate of H2 to disappear is 2.0 moles per second, what is the rate of appearance of NH3 at the same time? N2(g) 3
nasty-shy [4]

Answer:

1.33 moles per second

Explanation:

From the balanced equation of reaction, 3 moles of H2(g) need to disappear/be consumed before 2 moles of NH3(g) can be formed/appear.

<em>According to reaction law, the rate of disappearance of reactants is equal to the rate of appearance of products.</em>

Now, only 2 moles per seconds of H2 has disappeared initially;

3 moles H2 is needed for 2 moles NH3

2 moles H2 will give;        2 x 2/3 = 4/3 or 1.33 moles

<em>Hence, if the initial rate of disappearance of H2 is 2.0 moles per second, the initial rate of appearance of NH3 would be </em><em>1.33 moles per second.</em>

4 0
3 years ago
N the following reaction, how many grams of nitroglycerin C3H5(NO3)3 will decompose to give 120 grams of water?
creativ13 [48]
4 C₃H₅(NO₃)₃(l) = 12 CO₂(g) + 6 N₂<span>(g) + 10 H</span>₂O(g) + O₂<span>(g)

4 x ( 227.0995 ) g  ---------------> 10 x ( 18.0158) g
mass of </span><span>nitroglycerin-------------> 120 g

</span>m ( nitroglycerin ) = 120 x 4 x 227.0995 / 10 x <span>18.0158
</span>
m ( <span>nitroglycerin ) = 109007.76 / 180.158

m ( </span>nitroglycerin ) = 605.06 g

hope this helps!
3 0
4 years ago
Read 2 more answers
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
4 years ago
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