<h3>
Answer:</h3>
4 cm³
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtract Property of Equality
<u>Chemistry</u>
<u>Gas Laws</u>
Density = Mass over Volume
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
D = 25 g/cm³
m = 100 g
<u>Step 2: Solve for </u><em><u>V</u></em>
- Substitute variables [D]:
- Multiply <em>V</em> on both sides:
- Isolate <em>V</em>:
Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.
Work:
1) Unbalanced chemical equation (given):
<span>Co + AgNO3 → Co(NO3)2 + Ag
2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag
3) mole ratios
1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag
4) Convert the masses in grams of the reactants into number of moles
4.1) 5.85 grams of Co
# moles = mass in grams / atomic mass
atomic mass of Co = 58.933 g/mol
# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol
4.2) 15.8 grams of Ag(NO3)
# moles Ag(NO3) = mass in grams / molar mass
molar mass AgNO3 = 169.87 g/mol
# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol
5) Limiting reactant
Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.
That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.
6) Product formed.
Use this proportion:
2 mol Ag(NO3) 0.0930mol Ag(NO3)
--------------------- = ---------------------------
2 mol Ag x
=> x = 0.0930 mol
Convert 0.0930 mol Ag to grams:
mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g
Answer 1: 10.03 g of siver metal can be formed.
6) Excess reactant left over
1 mol Co x
----------------------- = ----------------------------
2 mole Ag(NO3) 0.0930 mol Ag(NO3)
=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted
Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol
Convert to grams:
0.0528 mol * 58.933 g/mol = 3.11 g
Answer 2: 3.11 g of Co are left over.
</span>
Answer:
Do you mean a biology lab
Explanation:
what's the question
Answer:
See Explanation
Explanation:
Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.
2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4
2^1 = 2^1
The rate of reaction is first order with respect to Hbn
Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.
1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4
3^1 = 3^1
The reaction is also first order with respect to CO
b) The overall order of reaction is 1 + 1=2
c) The rate equation is;
Rate = k [CO] [Hbn]
d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4/6.7 * 10^-7
k = 4.7 * 10^2 mmol-1 L s-1
e) The reaction occurs in one step because;
1) The rate law agrees with the experimental data.
2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.
Answer: -
- 0.348°C is the melting point of a solution in which 2.5 grams of sodium chloride is added to 230 mL of water
Explanation: -
Mass of NaCl = 2.5 g
Molar mass of NaCl = 23 x 1 + 35.5 x 1 = 58.5 g / mol
Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
= 2.5 /58.5
= 0.043 mol
230mL of water has 0.043 mol.
1000mL of water has (0.043 mol x 1000 mL ) / 230mL
= 0.187 mol
Thus molality of NaCl =0.187 m
The molal freezing point depression constant of water = 1.86 °C/m
Depression in freezing point = 1.86 °C/m x 0.187 m
= 0.348°C
Melting point = 0 - 0.348°C
= - 0.348°C
