Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
Moles are an estimation of the smallest unit of the molecules and the atoms in a sample. The moles of ammonium nitrate in a sample are 0.5010 moles.
<h3>What are moles?</h3>
Moles are calculated by dividing the mass of the substance in gm by that of the molar mass in gram per mole.
Given.
Mass of ammonium nitrate = 40.10 gm
The molar mass of ammonium nitrate = 80. 0432 g/mol
Moles of ammonium nitrate are calculated as:
Therefore, moles of ammonium nitrate present is option d. 0.5010 moles.
Learn more about moles here:
brainly.com/question/2396149
Hey, for #1, it's 1, 1, 1 because it's already balanced!
For #2, it's 1, 2, 1, 1 since the product side has 2 Cl's but left side only had 1.
1) State the balanced chemical equation
Na3 PO4 + 3 AgNO3 → 3NaNO3 + Ag3 PO4
sodium phosphate silver nitrate sodium nitrate silver phosphate
2) State the molar ratios
1 mol Na3PO4 : 3 mol AgNO3 : 3 mol NaNO3 : 1 mol Ag3 PO4
3) As you see 3 moles of Silver Nitrate react with 1 mol of Na3 PO4, then you will need.
The you need to use the molar ratio 1:3 to calculate the number of moles of sodium phosphate
1 mol Ag NO3 * [ 1 mol Na3 PO4 / 3 mol Ag NO3] = 0.33 mol Na3 PO4
Answer: 0.33 mol sodium phosphate
Answer:
True
Explanation:
Fe³⁺ is an oxidizing agent with E° = +0.77V. Zn metal is a reducing agent with E° = -0.76V. This is a "downhill" reaction, where the oxidizing agent has a higher E° than the reducing agent. Therefore, the reaction is spontaneous.
