From ideal gas equation that is PV=nRT
n(number of moles)=PV/RT
P=760 torr
V=4.50L
R(gas constant =62.363667torr/l/mol
T=273 +273=298k
n is therefore (760torr x4.50L) /62.36367 torr/L/mol x298k =0.184moles
the molar mass of NO2 is 46 therefore density= 0.184 x 46=8.464g/l
Answer: The volume of 0.235 M 
needed to titrate 40.0 mL of 0.0500 M 
is 8.51 ml
Explanation:
To calculate the volume 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
We are given:
Putting values in above equation, we get:
Thus volume of 0.235 M needed to titrate 40.0 mL of 0.0500 M is 8.51 ml
Answer:
d ≈ 0.098 g/mL
Explanation:
The density of a substance can be found by dividing the mass by the volume.
d=m/v
The mass of the substance is 0.221 grams and the volume is 2.25 milliliters.
m= 0.221 g
v= 2.25 mL
Substitute the values into the formula.
d= 0.221 g / 2.25 mL
Divide
d= 0.098222222 g/mL
Let’s round to the nearest thousandth. The 2 in the ten thousandths tells us to keep the 8 in the thousandth place.
d ≈ 0.098 g/mL
The density of the substance is about 0.098 grams per milliliter.
It is Cobalt(II)chloride dihydrate. The value of x is 2.
First, we have to determine the mass of H₂O that was evaporated upon heating which will be = 2.00g - 1.56 g
= 0.435 g
So,
CoCl₂ + x.H₂O → CoCl₂.xH₂O
Molecular mass of CoCl₂ = 129.839 g/mol
Molecular mass of H₂O = 18 g/mol
Weight of sample after heating = 1.565 g
Weight of H₂O evaporated = 0.435 g
Hence,
x ≈ 2
So, it is Cobalt(II)chloride dihydrate.
Learn more about molecular mass here:
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Answer:
The approximate molar mass of NaOH is 40 g/mol. You have 4 g NaOH.
No. of moles of NaOH is given by
n=(Given mass)/(Molar mass)
= 4/40 = 0.1 mol
1 mol has 6.022 × 10^23 molecules.
Thus, 0.1 mol has 6.022 × 10^23 × 0.1 molecules i.e. 6.022 × 10^22 molecules.
Explanation:
