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Vsevolod [243]
3 years ago
6

Convert 4.6 atm to mmHg.

Chemistry
1 answer:
Evgen [1.6K]3 years ago
3 0
It’s
3496 you can google this you know
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Based on the electron configuration of the two
Whitepunk [10]

Answer:

A

Explanation:

Cuz the when you pair Li and Cl, it becomes LiCl (it has 1:1 ratio)

4 0
3 years ago
Read 2 more answers
The mixing of which pair of reactants will result in a precipitation reaction?
Maru [420]

Answer:

K2SO4(aq) + Ba(NO3)2(aq)

Explanation:

K2SO4(aq) + Ba(NO3)2(aq)= 2KNO3(aq) + BaSO4(s)

The reaction produces BaSO4

Which precipitates as the insoluble product and Soluble KNO3 solution

8 0
3 years ago
Which of the following is not a correct chemical equation for a double displacement reaction?
Step2247 [10]

Answer:

B. CaCl + LiCO3 yields CaCO3 + LiCl is not correct

It should be CaCl2 + Li2CO3  → 2LiCl + CaCO3

Explanation:

For a reaction to be double displacement reaction there are two things we need to look for

1) There must be an interchange of the group of ions

2) The reactants must dissolve in water to release ions

A. 2RbNO3 + BeF2 yields Be(NO3)2 + 2RbF

2Rb+ + NO3- + Be^2+ + 2F- → Be(NO₃)₂ + 2RbF

This is correct

B. CaCl + LiCO3 yields CaCO3 + LiCl

This is not correct

The correct equation is:

CaCl2 + Li2CO3  → Ca2+ + 2Cl- + 2Li+ + CO3^2-  → 2LiCl + CaCO3

C. Na3PO4 + 3KOH yields 3NaOH + K3PO4

3Na+ + PO4^3- + 3K+ + 3OH- → 3NaOH + K3PO4

This is correct

D. 2MgI2 + Mn(SO3)2 yields 2MgSO3 + MnI4

2Mg^2+ + 4I- + Mn^4+ + 2SO3^2- → 2 MgSO3 + MnI4

This is correct

8 0
3 years ago
6
Anestetic [448]

when you type your answer ... your body gets sick rid of waste through the skin

8 0
3 years ago
Could someone help me with this?
Varvara68 [4.7K]

Solving part-1 only

#1

KMnO_4

  • Transition metal is Manganese (Mn)

#2

Actually it's the oxidation number of Mn

Let's find how?

\\ \tt\Rrightarrow x+1+4(-2)=0

\\ \tt\Rrightarrow x+1-8=0

\\ \tt\Rrightarrow x-7=0

\\ \tt\Rrightarrow x=+7

  • x is the oxidation number

#3

  • Purple as per the color of potassium permanganate

#4

\boxed{\begin{array}{c|c|c}\boxed{\bf Tube} &\boxed{\bf Charge} &\boxed{\bf No\:of\; electrons\: loss}\\ \sf 2 &\sf +6 &\sf 6e^-\\ \sf 3& \sf +2 &\sf 2e- \\ \sf 4 &\sf 4 &\sf 4e^-\end{array}}

7 0
2 years ago
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