Just divide the two (2 / 0.05) and you will get your answer; there are 40 drops of bloodin the collection tube.
Answer:
10.28 mol
Explanation:
S + 2O = SO2
(atm x L) ÷ (0.0821 x K)
(3.45 x 45.6) ÷ (0.0821 x 373)
=5.13726
Then round it to significant figures
=5.14
5.14 mol SO2 x (2 mol O ÷ 1 mol SO2)
=10.28 mol O
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Answer: hello your question is incomplete below is the complete question
Salt water contains n sodium ions (Na+) per cubic meter and n chloride ions (Cl−) per cubic meter. A battery is connected to metal rods that dip into a narrow pipe full of salt water. The cross sectional area of the pipe is A. The magnitude of the drift velocity of the sodium ions is VNa and the magnitude of the drift velocity of the chloride ions is VCl.
What is the magnitude of the ammeter reading ?
answer :
| I | = neAVₙₐ + neAV (Cl-)
Explanation:
Given that there are N sodium ions
<u>Determine the Magnitude of the ammeter reading </u>
| I | = current due to sodium ions + current due to (Cl-) ions
= neAVₙₐ + neAV (Cl-)
