Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
The statement that describes a chain reaction brought about by a nuclear reaction is "neutrons <span>released during a fission reaction cause other nuclei to split." This is applicable to fission reactions only wherein atoms split and produce neutrons that also cause further atoms to split, thus creating a chain or series of reactions.</span>
86 percent is the percent yield for this experiment if he expected to produce 5g of product.
Explanation:
Given that:
mass of test tube = 5 grams
mass of test tube + reactant is 12.5 grams
mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)
mass of reactant = 12.5 -5
= 7.5 grams
when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.
so mass of product formed = 9.3 - 5
= 4. 3 grams of product is formed (actual yield)
However, he expected the product to be 5 grams (theoretical yield)
Percent yield =
x 100
putting the values in the formula:
percent yield =
x 100
= 86 %
86 percent is the percent yield.
Answer:
5 000 000 (5 million atoms)
Explanation:
Let us assume that a vanadium atom has a spherical shape.
diameter of a sphere = 2 x radius of the sphere
Thus,
Radius of a vanadium atom = 130 pm
= 130 x
m
The diameter of a vanadium atom = 2 x radius
= 2 x 130 x
= 260 x
m
Given a distance of 1.30 mm = 1.30 x
m,
The number of vanadium atoms required to span the distance = 
= 5000000
Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.