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2 years ago

14

BRAINLIESTTT ASAP! PLEASE HELP ME :)

1 answer:

Mashcka [7]2 years ago

6 0

London dispersion forces I believe.

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Please help me thanks:)

lianna [129]

Answer:

where it goes

Explanation:

3 0

2 years ago

Read 2 more answers

What did J.J. Thomson discover about the atom that changed the atomic model previously used?

galina1969 [7]

J.J. Thomson hypothesized and discovered that the atom was not the smallest unit of matter but that instead there were much smaller units. He discovered "sub-atomic particles" which make up atoms. The sub-atomic particle that Thomson discovered was the electron. He discovered this through a process of experiments testing cathode rays.

7 0

2 years ago

How moles are in 45.7g of CH3?

miskamm [114]

first find the atomic weight of CH3 which would be

atomic weight: 12.011 (3×1.008) = 36.32 g/mol

then find the moles in the given mass

36.32 ÷ 45.7 = 0.794

I HOPE I'M NOT WRONG I HAVENT DONE CHEM IN SO LONG

6 0

2 years ago

A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

so

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

8 0

2 years ago

The solubility product of AgCl is 1.4 x 10-4 at 100°C. Calculate the solubility of AgCl in

marshall27 [118]

Answer:

AgCl=Ag+ + Cl-

1.4×10^-4 =x × x

x²=1.4×10^-4

x=√(1.4×10^-4)

x=0.012

5 0

2 years ago