Calcium citrate malate contains the calcium salt of citric acid and malic acid
<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
Answer: Climax Commuity.
Explanation: (hope it’s right! Good luck!)
Answer:
well since 3 is greater than 2 it would be 3 moles of sulfur.
Explanation:
Answer:
6. 7870 kg/m³ (3 s.f.)
7. 33.4 g (3 s.f.)
8. 12600 kg/m³ (3 s.f.)
Explanation:
6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.
1 kg= 1000g
1m³= 1 ×10⁶ cm³
Mass of iron bar
= 64.2g
= 64.2 ÷1000 kg
= 0.0642 kg
Volume of iron bar
= 8.16 cm³
= 8.16 ÷ 10⁶
Density of iron bar
= 7870 kg/m³ (3 s.f.)
7.
Mass
= 1.16 ×28.8
= 33.408 g
= 33.4 g (3 s.f.)
8. Volume of brick
= 12 cm³
Mass of brick
= 151 g
= 151 ÷ 1000 kg
= 0.151 kg
Density of brick
= mass ÷ volume
(3 s.f.)
