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S_A_V [24]
3 years ago
9

Plz help me with this plz pt.2

Chemistry
1 answer:
bogdanovich [222]3 years ago
3 0
The answer to your question is A I think
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What is the oxidation number of phosphorus (P) in sodium phosphate (Na3PO4)?
joja [24]
Na⁺¹₃P⁺⁵O⁻²₄

+1*3 + (+5) + (-2*4) = 0
8 0
3 years ago
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Benzene has a specific gravity of 0.88. if it was spilled in a river what would it do?
Gelneren [198K]
As we know,
                    Density of Benzene  =  876 Kg/m³
And,
                    Density of Water       =  997 Kg/m³
So,
     Specific Gravity is calculated as,

                     Specific Gravity  =  Density of Benzene / Density of Water

                     Specific Gravity    =   876 Kg/m³ / 997 Kg/m³

                     Specific Gravity    =  0.878

Every object having specific gravity less than 1 will float on water and if value is greater than 1 then it will sink.

Benzene being non-polar in nature does not mix with water and due to less density it will float on the surface of water.
4 0
3 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
Question 4 of 10
frez [133]

Answer: D. Slow down the chain reaction by absorbing free neutrons

Explanation: just got it right on the quiz A P E X

3 0
2 years ago
A shampoo has a pH of 8.59. What is the [H3O+] in the shampoo?
Cerrena [4.2K]

Answer:

2.57 e-9

Explanation:

The formula is H3O=10^-Ph

10^-8.59=2.57 e-9

6 0
2 years ago
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