Answer:
V = 5 cm³
ρ = 4 g/cm³
Explanation:
Step 1: Calculate the volume (V)
We have a wooden cuboid of dimensions 5 cm × 1 cm × 1 cm. We can calculate its volume using the following expression.
V = 5 cm × 1 cm × 1 cm
V = 5 cm³
Step 2: Calculate the density (ρ)
The density is equal to the mass divided by the volume.
ρ = m / V
ρ = 20 g / 5 cm³
ρ = 4 g/cm³
Answer:
69.7% is percent yield
Explanation:
Based on the reaction:
3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)
2 moles of Na3PO4 react producing 6 moles of NaNO3.
As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:
24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =
72.6 moles of NaNO3
As there are produced 50.6 moles of NaNO3, percent yield is:
50.6 moles NaNO3 / 72.6 moles NaNO3 =
<h3>69.7% is percent yield</h3>
Answer:
0.725 kg
Explanation:
Step 1: Given data
- Kinetic energy of the softball (K): 145 J
- Speed of the softball (v): 20.0 m/s
- Mass of the softball (m): ?
Step 2: Calculate the mass of the softball
We will use the following expression.
K = 1/2 × m × v²
m = 2 K / v²
m = 2 × 145 J / (20.0 m/s)²
m = 0.725 kg
The mass of the softball is 0.725 kg.
Answer: N = 2.78 × 10^23 atoms
There are N = 2.78 × 10^23 atoms in 70g of Au2cl6
Completed Question:
Calculate the number of gold atoms in a 70g sample of gold(III) chloride . Be sure your answer has a unit symbol if necessary, and round it to significant digits
Explanation:
Given:
Molar mass of Au2cl6 = 303.33g/mol
Mass of Au2cl6 = 70g
Number of moles of Au2cl6 = 70g/303.33g/mol = 0.231mol
According to the chemical formula of Au2cl6,
1 mole of Au2cl6 contains 2 moles of Au
Number of moles of Au = 2 × 0.231mol = 0.462mole
There are 6.022 × 10^23 atoms in 1 mole of an element.
Number of Atom of gold in 0.462 mole of gold is:
N = 0.462 mol × 6.022 × 10^23 atoms/mol
N = 2.78 × 10^23 atoms
False, energy conversion just means the energy is going to be used by another force
