KE = mv²/2
m=2*KE/v²
v=50 m/s
KE=500J
m=2*KE/v² =2*500/50²=1000/2500= 0.4 kg
Answer:
The equation for molarity is moles/liter for the first question you would do 0.256/0.143 liters to get 1.790 mol/L
Explanation:
The second problem you would do need to find the moles of NaCl which you would do by doing 4.89 g/58.44g/mol= 0.08367 then do 0.08367/0.600= 0.139 mol/L
The third problem would be the same steps as the second one.
The fourth problem would be (0.460M)(5.50L)= 2.53 moles
#AB
Electronegativity difference=3.3-2.9=0.4.
- It's a covalent bond.
- Gaseous or solid substance.
#AC
Electronegativity difference=3.3-0.7=2.6
- Its an ionic bond.
- Solid substance.
#BC
Electronegativity difference=2.9-0.7=2.3
- It's an ionic bond
- Solid substance
There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
4 moles KOH ------------------- moles Mg(OH)₂
moles Mg(OH)₂ = 4 x 1 / 2
= 2 moles of Mg(OH)₂
molar mass Mg(OH)₂ = 58g/mol
mass of Mg(OH)₂ = n x mm
mass of Mg(OH)₂ = 58 x 2
= 116 g of <span> Mg(OH)₂</span>
hope this helps!
