Answer:
5.9x10²² atoms of Cu are in one naira coin
Explanation:
To solve this question we need to find the mass of Copper. Then, using its molar mass (Cu = 63.546g/mol) we must find the moles of Cu and its atoms using Avogadro's number:
<em>Mass Cu:</em>
7.3g * 86% = 6.278g is the mass of Cu.
<em>Moles Cu:</em>
6.278g * (1mol / 63.546g) = 0.099moles Cu
<em>Atoms Cu:</em>
0.099moles Cu * (6.022x10²³atoms / 1mol) =
<h3>5.9x10²² atoms of Cu are in one naira coin</h3>
Answer:
(D.)
The core of the pyruvate dehydrogenase complex is made up of eight catalytic <u>trimers </u>that make up the<u> E2</u> component.
Explanation:
Eight trimers assemble as a hollow truncated cube, which forms the core of the multi-enzyme complex, known as the E2 complex in human pyruvate dehydrogenase complex.
1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.
Let's consider the reaction for the combustion of Mg.
Mg + 1/2 O₂ ⇒ MgO
1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:
We can calculate the mass percent of O in MgO using the following expression.
You can learn more about mass percent here: brainly.com/question/14990953
Explanation:
d. endothermic change as
heat is added to solid ice to change it to liquid water
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
