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Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+
From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3
Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3
The answer is 2, liquid to vapor because vaporization is the process of liquids to vapors.
Answer: 670K
Explanation:
Given that,
Original volume of gas V1 = 1.22 L
Original temperature T1 = 286 K
New volume V2 = 2.86 L
New temperature T2 = ?
Since volume and temperature are involved while pressure is constant, apply the formula for Charles law
V1/T1 = V2/T2
1.22 L/286 K = 2.86 L/ T2
Cross multiply
1.22 L x T2 = 286 K x 2.86 L
1.22T2 = 817.96
Divide both sides by 1.22
1.22T2/1.22 = 817.96/1.22
T2 = 670.459 K (Round to the nearest whole number as 670 K)
Thus, the temperature of the gas is 670 Kelvin
<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>
