Answer:
1.0 M is the concentration of hydrochloric acid.
Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:


1.0 M is the concentration of hydrochloric acid.
A) cesium chloride
B) barium oxide
C) potassium sulfide
D) beryllium chloride
E) hydrogen bromide
F) aluminum fluoride
Answer:
463.0 g.
Explanation:
- We can use the following relation:
<em>n = mass/molar mass.</em>
where, n is the mass of copper(ii) fluoride (m = 4.56 mol),
mass of copper(ii) fluoride = ??? g.
molar mass of copper(ii) fluoride = 101.543 g/mol.
∴ mass of copper(ii) fluoride = (n)(molar mass) = (4.56 mol)(101.543 g/mol) = 463.0 g.
Answer:
a
The structure of the expected product for this reaction is shown on the first uploaded image
b
The structure of the expected product for this reaction is shown on the first uploaded image
c
The structure of the expected product for this reaction is shown on the first uploaded image
d
The structure of the expected product for this reaction is shown on the first uploaded image
e
The structure of the expected product for this reaction is shown on the first uploaded image
Explanation:
The Explanation is shown on the second third and fourth uploaded image
For a Forty miles above the earth's surface, the temperature at 250k and the pressure at only 0.20 torr, the density of air at this altitude is mathematically given as
d=3.732*10{-4}g/l
<h3>What is the density of air at this altitude?</h3>
Generally, the equation for the ideal gas is mathematically given as
PV=(m/mw)RT
Therefore
d=m/v
d=26,664*20/83.14*250
d=0.37152
d=3.732*10{-4}g/l
In conclusion, the density
d=3.732*10{-4}g/l
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