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Alex
3 years ago
7

Which of the following substances is considered to be a base? A. HCl B. HNO3 C. H2SO4 D. Ca(OH)2

Chemistry
2 answers:
Ganezh [65]3 years ago
6 0

I took this quiz and picked D and got it correct

PtichkaEL [24]3 years ago
3 0
Among all of the choices listed, the only option that is a base would be the last one. As it is unique, whereby Hydrogen is not present in the front of the compound.

Bases also possess OH in their molecular formula.
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A technician measures the weight of a reagent . The weight is 3269 g. What is the weight in kilograms ? Write your answer as a d
Georgia [21]

Answer:

3.269 kg

Explanation:

1000g = 1 kg. Since kg is the larger unit, you need to divide by 1000 to get kilograms. You can do the problem as a unit conversion:

3269 g × <u>1 kg</u>  =  3.269 kg

                1000g

Hope this is helpful!

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7 0
3 years ago
What is the salt that is produced when calcium hydroxide (Ca(OH)2) reacts with sulfuric acid (H2SO4)? CaSH2Ca H2O CaSO4
Wittaler [7]
H2SO4+CA[OH]2=CASO4+2H2O
3 0
3 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
3 years ago
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