6+2=115 and its good it took test
Answer:
Mass = 114.26 g
Explanation:
Given data:
Number of gold atoms = 3.47×10²³ atoms
Mass in gram = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
3.47×10²³ atoms × 1 mol /6.022 × 10²³ atoms
0.58 mol
Mass of gold:
Mass = number of moles × molar mass
Mass = 0.58 mol × 197 g/mol
Mass = 114.26 g
Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
Answer:
Explanation:
Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)
We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products
(a) Enthalpies of formation of reactants and products
(b) Total enthalpies of reactants and products
(c) Enthalpy of reaction
Answer: With few exceptions, the mitotic process ensures that this is the case. Therefore, mitosis ensures that each successive cellular generation has the same genetic composition as the previous generation, as well as an identical chromosome set.
