To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
2 M x V1 = 0.1 M x .5 L
<span>V1 = 0.025 L or 25 mL of the
2 M KCl solution is needed</span>
(a) Reaction of nitric acid with non-metal:
C+4HNO
3
⟶CO
2
+2H
2
O+4NO
2
S+6HNO
3
⟶H
2
SO
4
+2H
2
O+6NO
2
(b) Nitric acid showing acidic character:
K
2
O+2HNO
3
⟶2KNO
3
+H
2
O
ZnO+2HNO
3
⟶Zn(NO
3
)
2
+H
2
O
(c) Nitric acid acting as oxidizing agent
P
4
+20HNO
3
⟶4H
3
PO
4
+4H
2
O+20NO
2
3Zn+8HNO
3
⟶3Zn(NO
3
)
2
+4H
2
O+2NO
hope that helps you please mark brainliest
I can’t see the choices can you take another picture of this assignment please so i can help you
Answer:
A tritium is produced.
Explanation:
Combining two additional neutrons to the nucleus of the hydrogen atom makes it a tritium, Hydrogen-3.
neutron is designated ¹₀n; this shows a mass number of 1 and no atomic number
Hydrogen-1 is designated as ₁¹H; a mass number of 1 and atomic number of 1. This particle is actually more like a proton.
Combining both:
₁¹H + 2¹₀n → ³₁H
This is a nuclear reaction and in balancing such reaction equation, mass numbers and atomic numbers must be conserved.
<u>Answer:</u> The atomic mass of these species is different and atomic number remains same.
<u>Explanation:</u>
Isotopes are the chemical species of the same element having different number of neutrons.
- Atomic number is equal to the number of protons or electrons present in that element.
Atomic Number = Number of electrons = Number of protons
- Atomic mass is defined as the sum of number of protons and neutrons contained in an atom.
Atomic Mass = Number of protons + Number of neutrons
For isotopes, as the number of neutrons differ, the atomic mass also differs.
For Example: Carbon has 3 naturally occurring isotopes: 
. The atomic number remains the same but atomic mass differs.
Hence, for isotopes, the atomic mass of these species is different and atomic number remains same.
