Answer:
im pretty sure 2+
Explanation:
subtract the electrons from the protons
Answer: The statement is true
Explanation:
The half-life of a radioactive isotope is the time taken for half of the total number of atoms in a given sample of the isotope to decay.
For instance
The half-life of radium is 1622 years. This means that if we have 1000 radium atoms at the beginning, then at the end of 1622 years, 500 atoms would have disintegrated, leaving 500 undecayed radium atoms
Thus, the statement is true
Hydrogen gas and oxygen gas react to form liquid water according to the following equation:
2H₂ + O₂ → 2H₂O
a. Converting our given masses of each gas to moles, we have:
(25 g H2)/(2 × 1.008 g/mol) = 12.4 mol H2; and
(25 g O2)/(2 × 15.999 g/mol) = 0.781 mol O2.
From the equation, two moles of H2 react with every one mole of O2. To fully react with 12.4 moles of H2, as we have here, one would need 6.2 moles of O2, which is far more than what we're actually given. Thus, the oxygen is our limiting reactant, and as such it will be the first reactant to run out.
b. Since O2 is our limiting reactant, we use it for determining how much product, in this case, H2O, is produced. From the equation, there is a 1:1 molar ratio between O2 and H2O. Thus, the number of moles of H2O produced will be the same as the number of moles of O2 that react: 0.781 moles of H2O. The mass of water produced would be (0.781 mol H2O)(18.015 g/mol) ≈ 14 grams of water (the answer is given to two significant figures).
c. Since the hydrogen reacts with the oxygen in a 2:1 ratio, twice the number of moles of oxygen in hydrogen is consumed: 0.781 mol O2 × 2 = 1.562 mol H2. Since we began with 12.4 moles of H2, the remaining amount of excess H2 would be 12.4 - 1.562 = 10.838 mol H2. The mass of the excess hydrogen reactant would thus be (10.838 mol H2)(2 × 1.008 g/mol) ≈ 22 grams of hydrogen gas (the answer is given to two significant figures).
Answer:
101.56 of H₂O
Explanation:
The balanced equation for the reaction is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Next, we shall determine the mass of CH₄ that reacted and the mass of H₂O produced from the balanced equation. This is illustrated below:
Molar mass of CH₄ = 12 + (4×1.01)
= 12 + 4.04
= 16.04 g/mol
Mass of CH₄ from the balanced equation = 1 × 16.04 = 16.04 g
Molar mass of H₂O = (2×1.01) + 16
= 2.02 + 16
= 18.02 g/mol
Mass of H₂O from the balanced equation = 2 × 18.02 = 36.04g
SUMMARY:
From the balanced equation above,
16.04 g of CH₄ reacted to produce 36.04 g of H₂O.
Finally, we shall determine the mass of water, H₂O produced by the reaction of 45.2 g of methane, CH₄. This can be obtained as illustrated below:
From the balanced equation above,
16.04 g of CH₄ reacted to produce 36.04 g of H₂O.
Therefore 45.2 g of CH₄ will react to produce = (45.2 × 36.04)/16.04 = 101.56 g of H₂O.
Thus, 101.56 of H₂O were obtained.
Answer:
[CICH2COOH] = 0.089 M
[CICH2COO-] = 0.011 M
[H+] = 0.011 M
Ka = 1.36 x 10^-3
Explanation:
The reaction
CICH2COOH ⇄ H+ (aq) + CICH2COO- (aq)
The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.
let us calculate Ka
First , find change in concentration
since , 11% ionized
change in concentration = 0.10 X 11% = 0.011 M
Initial Concentration of CICH2COOH = 0.10 M
change in concentration of CICH2COOH = - 0.011 M
Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m
Initial Concentration of CICH2COO- = 0 M
change in concentration of CICH2COO- = + 0.011 M
Equilibrium Concentration of CICH2COO- = 0.011 M
Initial Concentration of H+ = 0 M
change in concentration of H+ = + 0.011 M
Equilibrium Concentration of H+ = 0.011 M
Therefore,
[CICH2COOH] = 0.089 M
[CICH2COO-] = 0.011 M
[H+] = 0.011 M
Ka = [H+][CICH2COO-] /[CICH2COOH]
Ka = (0.011 * 0.011) / (0.089)
Ka = 1.36 x 10^-3
