<span>Answer:
Pb(NO3)2+2HCl=PbCl2+2HNO3
Step 2
Find # of moles of Pb
13.87g Pb(NO3)2 x 1 mol/331.22=.041875 mol Pb(NO3)2
Step 3
Use balanced equation to find weight of Pb(NO3)2
.041875mol Pb x 1 mol Cl2/1mol Pb x 70.90g/1 mol Cl2=2.9689g Cl2
.041875 mol Pb x 207.2/1mol Pb=8.6765g Pb
=11.65g</span>
<span>2<span>C6</span><span>H6</span>O(l)+17<span>O2</span>−−>12C<span>O2</span>(g)+12<span>H2</span>O(l)</span>
<span>2S<span>O2</span>(g)+<span>O2</span>(g)−−>2S<span>O3</span>(g)</span>
<span><span>N2</span>(g)+<span>O2</span>(g)−−>2NO(g)</span>
<span>2Na(s)+B<span>r2</span>(l)−−>2NaBr(s)</span><span>
On the 1st 3 I have
12 -17 = -5
2 - 3 = -1
2 - 2 = 0
For the last one:
</span><span>Delta n=0</span><span>
</span>
Answer is: the pressure in a vessel is 1.48 atm.
V(Cl₂) = 22.4 L; pressure of chlorine gas.
n(Cl₂) = 1.50 mol; amount of chlorine gas.
T = 0.00°C = 273.15 K; temperature.
a = 6.49 L²·atm/mol²; the constant a provides a correction for the intermolecular forces.
b = 0.0562 L/mol; value is the volume of one mole of the chlorine gas.
R = 0.08206 L·atm/mol·K, universal gas constant.
Van de Waals equation: (P + an² / V²)(V - nb) = nRT.
(P + 6.49 L²·atm/mol² · (1.5 mol)² / (22.4 L)²) · (22.4 L - 1.5 mol·0.0562 L/mol) = 1.5 mol · 0.08206 L·atm/mol·K · 273.15 K.
(P + 6.49 L²·atm/mol² · (1.5 mol)² / (22.4 L)²) = (1.5 mol · 0.08206 L·atm/mol·K · 273.15 K) ÷ (22.4 L - 1.5 mol · 0.0562 L/mol).
P + 0.029 atm = 33.62 L·atm ÷ 22.31 L.
P = 1.507 atm - 0.029 atm.
P = 1.48 atm; the pressure.
Answer:
c. isotope number
Explanation:
Mass Number is the sum total of mass of protons and neutrons present in the nucleus of an atom. Generally they are being used in distinguishing isotopes. E.g Carbon - 12, Carbon - 13
Atomic Number is the number of protons. Every single element has it's unique atomic number and can be used in identification purpose. E.g Carbon - 6, Hydrogen - 1.
The correct option is option C. This is the symbol that is not necessary for the identification of a nuclide.