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ruslelena [56]
3 years ago
13

When hydrochloric acid reacts with sodium sulfide, hydrogen sulfide gas and sodium chloride are produced. How many moles of H2S

will be generated by 2.56 moles of HCl? 2HCl + Na2S → H2S + 2NaCl
A 10 liters
B 0.1 liters
C 40 liters
D 20 liters
Chemistry
1 answer:
lidiya [134]3 years ago
5 0
Well 1.28 mol of H2S will be produce but it didn't said anything about that where the experiment happens (in STP situation (which seems not to be true ) or not )
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The elements least likely to form bonds are found in which group?
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Answer:

Noble gases are a unique set of elements in the periodic table because they don't naturally bond with other elements.

Explanation:

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4 0
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
How many grams are there in 2.75E23 molecules of C2H6 (MM = 30 grams)?
skelet666 [1.2K]

Answer:13.70 i think

Explanation:

3 0
2 years ago
how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?
kobusy [5.1K]
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....


Do the calculation
6 0
3 years ago
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