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Nutka1998 [239]
3 years ago
12

PLEASE HELP ME ON THIS QUESTION

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
4 0
He call me big purr
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If you try to balance an equation by changing subscripts you change...
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Answer:

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Explanation:

sklwlrlfclxoskkekrdododosoekekrkrododowoekekfkdodkwkeororkdkdkwejrjrkfidiwi3jr

8 0
3 years ago
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pochemuha

Answer:

boiling point

Explanation:

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3 years ago
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Calculate the energy(J) change associated with an electron transition from n=2 to n=5 in a Bohr hydrogen atom. Put answer in sci
Oliga [24]
The energy at n level of hydrogen atom energy level =13.6/n^2

substiture the respective n values in the equation above and find the difference in the energy levels

instagram : imrajsingh

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3 years ago
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Bi2(CO3)3<br> Name of this compound
Keith_Richards [23]

Answer:

The name of this compound is :

Bi2(CO3)3 = Bismuth Carbonate

Explanation:

The name of the compound is derived from the name of the elements present  in it.

The rule followed while naming the compound are:

1. The first element (always the cation) is named as such .

2. The second element (The anion) end with "-ate ,  -ide ," etc

3. NO prefix is added while naming the first element.

For example : Bi2 can't be named as Dibismuth

Na2 = Can't be named as disodium

Hence the compound :

Bi2(CO3)3 contain two element : Bi and CO3. Here , Bi = cation (named as such) and CO3 = anion (named according to rules)

Bi = Bismuth

CO3 = carbonate

Bi2(CO3)3 = Bismuth Carbonate

The molecular mass of this compound is :

Molecular mass = 2 (mass of Bi) + 3(mass of C) + 6(mass of O)

= 2 (208.98)+3(12.01)+6(15.99)

= 597.987 u

5 0
3 years ago
A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th
grandymaker [24]

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization =  37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

8 0
3 years ago
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