Answer:
1.22 L of carbon dioxide gas
Explanation:
The reaction that takes place is:
- CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
First we <u>determine which reactant is limiting</u>:
- Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃
- Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl
So HCl is the limiting reactant.
Now we calculate the moles of CO₂ produced:
- 0.05 mol HCl *
= 0.05 mol CO₂
Finally we use PV=nRT to <u>calculate the volume</u>:
- T = 25 °C ⇒ 25 + 273.16 = 298.16 K
1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Answer:
1.78 × 10⁹ μg
Explanation:
We have to convert 1.78 kg to μg.
Step 1: Convert 1.78 kilograms to grams
We will use the conversion factor 1 kg = 10³ g.
1.78 kg × 10³ g/1 kg = 1.78 × 10³ g
Step 2: Convert 1.78 × 10³ grams to micrograms
We will use the conversion factor 1 g = 10⁶ μg.
1.78 × 10³ g × 10⁶ μg/1 g = 1.78 × 10⁹ μg
1. The newspaper solution that would work best with inks is alcohol since alcohol can dissolve inks.
2. Water can be used to carry salts since salts are soluble in water.
3. Oils can be used to carry fatty acids since they're both hydrophobic substances.<span />
Answer is: 2) 117g.
2Na + Cl₂ → 2NaCl
Step 1: calculate amount of substance of sodium and chlorine.
n(Na) = m(Na)÷M(Na) = 46g ÷ 23 g/mol = 2 mol.
n(Cl₂) = m(Cl₂)÷M(Cl₂) = 71g ÷ 71 g/mol = 1 mol.
Step 2: calculate amount of substance and mass of sodium-chloride.
Because both sodium and chlorine react completely, we can use both n to compare with n of NaCl.
n(Na) : n(NaCl) = 2:2, 2 mol : n(NaCl) = 2:2
n(NaCl) = 2mol, m(NaCl) = 2mol ·5805 g/mol = 117 g.
Answer:
curium
−
243
,
252
/
99
Es,
251
/
98
Cf,
214
/
82
Pb
Explanation: Im not very good with this but here ya go!
