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DerKrebs [107]
3 years ago
11

I kinda need this question answered fast please and thank you :)

Chemistry
1 answer:
Aloiza [94]3 years ago
6 0
D. A chemical bond

A mutual attraction between the nuclei and electrons in two different atoms is called a chemical bond, therefore the answer is D
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You place 10 grams of a salt into water and want it to dissolve. All of the following will cause a salt to dissolve faster excep
klio [65]

boiling, salt water will separate the salt from the water through the physical change of water from a liquid to a gas, leaving you with just salt left over.

8 0
2 years ago
if a gas is held in 3.60 L, 298 K AND 800. mmHg, what is the new pressure of the gas if the volume us decreased by half and the
sleet_krkn [62]

Answer:

Explanation:

By Ideal Gas Law, P1*V1 / T1 = P2*V2 / T2

So new pressure = (P1*V1 / T1) / (V2 / T2)

= P1*V1*T2 / T1*V2

= 800*3.6*298 / 250*1.8

= 1907.2 mmHg

6 0
2 years ago
Read 2 more answers
A compound is 2% H, 32.7% S, and 65.3% O by mass. What is the subscript on the O in the empirical formula for this compound?
IceJOKER [234]
Since all of those percents add up to 100, you can just directly convert that to grams. So now you can use 2 grams H, 32.7 grams S, and 65.3 grams O. Use that info and convert that to moles for an answer of 2mol H, 1mol S, and 4mol O. In every empirical question you need to divide each quantity of moles by the lowest number. In this case, that number is one, so they stay the same, but it's important to remember that step. You're final chemical formula would be H2SO4 and the answer to your question would be that the subscript for oxygen is 4. Hope this helped! 
4 0
2 years ago
what is the concentration (molarity) of a solution where 149g of kcl is dissolved in enough water to get a total volume of the s
Anit [1.1K]
Molar mass KCl = 74.55 g/mol

Number of moles:

mass KC / molar mass

149 / 74.55 => 1.998 moles 

Volume in liters: 500 mL / 1000 => 0.5 L 

Therefore:

M = moles / volume

M = 1.998 / 0.5

M = 3.996 mol/L⁻¹
4 0
3 years ago
Calculate the frequency required to eject electrons from gold with a work function of 8.76 x 10-19 J
Savatey [412]
hf=W+\frac{mv^{2}}{2}\\\\
f=\frac{W+\frac{mv^{2}}{2}}{h}\\\\\

W=8,76*10^{-19}J\\
m=9,11*10^{-31}kg\\
v=3*10^{8}\frac{m}{s}\\
h=6,63*10^{-34}Js\\\\\\
f=\frac{8,76*10^{-19}J+\frac{9,11*10^{-31}kg*(3*10^{8}\frac{m}{s})^{2}}{2}}{6,63*10^{-34}Js}=7,5*10^{19}Hz
5 0
3 years ago
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