Answer:
Moderate reaction means those reactions which proceed with a measurable rates at the normal room temperature.
Answer:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.
94.6 g. You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.
We can use a version of the <em>dilution formula</em>
<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2
where
<em>m</em> represents the mass and
<em>C</em> represents the percent concentrations
We can rearrange the formula to get
<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)
<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %
<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %
∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g
Answer:
d) V = 91.3 L
Explanation:
Given data:
Volume of nitrogen = ?
Temperature = standard = 273.15 K
Pressure = standard = 1 atm
Number of atoms of nitrogen = 2.454×10²⁴ atoms
Solution:
First of all we will calculate the number of moles of nitrogen by using Avogadro number.
1 mole = 6.022×10²³ atoms
2.454×10²⁴ atoms × 1 mol / 6.022×10²³ atoms
0.407×10¹ mol
4.07 mol
Volume of nitrogen:
PV = nRT
1 atm × V = 4.07 mol ×0.0821 atm.L /mol.K ×273.15 K
V = 91.3 atm.L /1 atm
V = 91.3 L
The balanced equation is 2HgO --> 2Hg + O2
