Answer:
21.02moles of KBr
Explanation:
Parameters given:
Number of moles BaBr₂ = 10.51moles
Complete reaction equation:
BaBr₂ + K₂SO₄ → KBr + BaSO₄
Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:
The balanced equation is:
BaBr₂ + K₂SO₄ → 2KBr + BaSO₄
From the equation:
1 mole of BaBr₂ produces 2 moles of KBr
∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr
Answer:
255.6
Explanation:
If you have 12 gallons and get 21.3mpg,
-Multiply 21.3 by 12
-you can travel 255.6 miles before running out of gas.
-If you need to estimate, round up to 256 miles.
The temperature of vinegar does increase the rate of reaction according to the collision theory.
Hope this answers your question!
Answer;
-The charges of the positive and negative copper ions cancel each other out.
Explanation;
-A normal atom has a neutral charge with equal numbers of positive and negative particles.That means an atom with a neutral charge is one where the number of electrons is equal to the atomic number.
-Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion. If it gains one or more electrons, it now carries a net negative charge, and is thus anionic.
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
