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lapo4ka [179]
2 years ago
5

PLEASE HELP THESE ARE MY LAST 3 QUESTIONS LEFT AND IM HAVING TROUBLE WITH THEM

Chemistry
2 answers:
Helga [31]2 years ago
7 0
1).................
Mashcka [7]2 years ago
3 0

Answer:

1)

2)Hg2 3 po4 2

Mercury(I) Phosphate (Hg2)3(PO4)2 Molecular Weight

3)2C6H14 + 19O2 → 12CO2 + 14H2O + heat

Explanation:

i dont know the 1st one i ggoles the 2nd one and it gave me this, hope it helps

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The major difference between a 1s orbital and a 2s orbital is that
Neko [114]

Answer:

The 2s orbital is at a higher energy level.

Explanation:

1s and 2s are the sub-orbitals that are located in an atom. They are nearest to the nucleus and are found on the s sub-orbital. The difference between 1s and 2s is the difference in their level of energy. 1s has low energy as compared to 2s. 1s orbital has the lowest energy because it is located closed to the nucleus. 2s orbital has higher energy than 1s because it's orbit is larger than 1s.

5 0
4 years ago
What is the numbers to the C02
andriy [413]

Answer:

413.20 ppm

Explanation:

4 0
3 years ago
At constant temperature the pressure on a 6.0 L sample of a gas is reduced from 2.0 atm to 1.0 atm. What is the new volume of th
kotegsom [21]
P1V1=P2V2, so P1V1/P2=V2.
2atm x 6.0 L/1.0 atm = 12.0 L
The new volume would be 12.0 Liters
4 0
3 years ago
Read 2 more answers
What is the correct name for Au3N?
Reptile [31]

Answer:

option D= Gold (I) nitride

Explanation:

The name of the given compound is gold(I) nitride.

Molar mass can be determine by following way:

molar mass Au3N = (molar mass of gold × 3) + (molar mass of nitrogen)

molar mass Au3N = (196.97 × 3 ) + ( 14 )

molar mass of Au3N =  590.91 g/mol + 14 g/mol

molar mass of Au3N = 604.91 g/mol

The nitrogen has valency of -3 so three Au(+1) will require while the valency of Au is (1+) one nitrogen will require to make the compound overall neutral.

Au3N

3(1+) + (-3) = 0

+3 - 3 = 0

0 = 0

The overall charge is 0, the compound will be neutral.

8 0
3 years ago
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s
Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 0.0632

t = time taken for the process  = ?

[A_o] = initial amount or concentration of the reactant  = 1.28 M

[A] = amount or concentration left time 't' = 1.28\times \frac{17}{100}=0.2176M

Now put all the given values in above equation, we get:

0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}

t=28s

Therefore, the time taken for the reaction is, 28 s.

8 0
3 years ago
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