<h3>Answer:</h3>
162.43 g of FeCl₂
<h3>
Explanation:</h3>
Step 1: Calculate mass of Fe;
As,
Density = Mass ÷ Volume
Or,
Mass = Density × Volume
Where Volume is the volume of water displaced = 10.4 mL
Putting values,
Mass = 7.86 g.mL⁻¹ × 10.4 mL
Mass = 81.744 g of Fe
Step 2: Calculate amount of FeCl₂;
The balance chemical equation is as follow,
Fe + 2 HCl → FeCl₂ + H₂ ↑
According to this equation,
55.85 g (1 mol) Fe produced = 110.98 g (1 mol) of FeCl₂
So,
81.744 g Fe will produce = X g of FeCl₂
Solving for X,
X = (81.744 g × 110.98 g) ÷ 55.85 g
X = 162.43 g of FeCl₂
Pan 4: theyre the smallest and most broken down :)
1) START counting for sig. figs. On the FIRST non-zero digit.
2) STOP counting for sig. figs. On the LAST non-zero digit.
3) Non-zero digits are ALWAYS significant.
4) Zeroes in between two non-zero digits are significant. All other zeroes are insignificant.
Answer:
0.313 mole of NH3
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2NH3(g) + 3CuO(s) → 3Cu(s) + N2(g) + 3H2O(g)
The number of mole of ammonia (NH3) required to react with 0.470 mole of copper(ll) oxide (CuO) can be obtained as follow:
From the balanced equation above,
2 moles of NH3 reacted with 3 moles of CuO.
Therefore, Xmol of NH3 will react with 0.470 mole of CuO i.e
Xmol of NH3 = (2 x 0.470) /3
Xmol of NH3 = 0.313 mole.
Therefore, 0.313 mole of NH3 is needed for the reaction
Answer:
It amplifies the picture of something far away.
Explanation:
