I think it is the third statement
‘A unit is made up of two or more atoms’
Answer:
The new pressure is 53.3 kPa
Explanation:
This problem can be solved by this law. when the volume remains constant, pressure changes directly proportional as the Aboslute T° is modified.
T° increase → Pressure increase
T° decrease → Pressure decrease
In this case, temperature was really decreased. So the pressure must be lower.
P₁ / T₁ = P₂ / T₂
80 kPa / 300K = P₂/200K
(80 kPa / 300K) . 200 K = P₂ → 53.3 kPa
A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is (
).
Utilize the titration method of in view that we're given the concentrations of every compound and the quantity of 
. Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.
- M1 = initial mass
- V1= initial volume
- M2 = final mass
- V2= final volume
- (0.138)(V1) = (0.205)x(26.0)
- V2=(0.205)x(26.0)\ 0.138
- V2 = 47.10 M/L
- The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L
Read more about the neutralization:
brainly.com/question/23008798
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Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
<h3>The volume of the sample is 17.4L</h3>
