The moles of any substance are equal to the substance's mass divided by its molar mass. Therefore, in order to calculate the moles of copper, you would divide the reacted mass by 63.55
Answer:
<h2>81.5 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volume
From the question we have
mass = 25 × 3.26
We have the final answer as
<h3>81.5 g</h3>
Hope this helps you
Answer:
A) 7.9 x 10⁶ inches
B) 1004 g
C) 2.8 x 10³ inches/ min
D) 1.2 x 10⁻⁴ mm
Explanation:
A) Since 39.37 inches = 1 m, you can convert meters to inches by multiplying by the conversion factor (39.37 inches / 1 m).
Notice that if 39.37 inches = 1 m then 39.37 inches / 1 m = 1. That means that when you multiply by a conversion factor, you are only changing units since it is the same as multiplying by 1 :
2.0 x 10⁵ m * (39.37 inches / 1 m) = 7.9 x 10⁶ inches
B) Conversion factors : (2.205 pounds / 1 kg) and (453.59 g / 1 pound), because 2.205 pounds = 1 kg and 1 pound = 453.59 g. Then:
1.004 kg * ( 2.205 pounds / 1 kg) * ( 453.59 g / 1 pound) = 1004 g
C) Conversion factor: (39.37 inches / 1 m) and (60 s / 1 min)
1.2 m/s * (39.37 inches / 1 m) * ( 60 s / 1 min) = 2.8 x 10³ inches/ min
D)Converison factor ( 1 mm / 1 x 10⁶ nm):
120 nm (1 mm / 1 x 10⁶ nm) = 1.2 x 10⁻⁴ mm
<u>Given:</u>
Enthalpy change (ΔH) for SO3 decomposition = +790 kJ
Moles of SO3 = 2.1 moles
<u>To determine:</u>
Energy required when 2.1 moles of SO3 reacts
<u>Explanation:</u>
The decomposition reaction is -
2SO3(g) → 2S(s) + 3O2 (g)
Energy required when 2 moles of SO3 reacts is 790 kJ
Thus, for 2.1 moles of SO3 the energy requirement would be
= 2.1 moles SO3 * 790 kJ/2 moles SO3 = 829.5 kJ
Ans: 830 kJ are required when 2.1 moles of SO3 reacts.
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
