Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>
<span>ΔT(freezing point)
= (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C
</span>
<span>
</span>
<span>Hope this answers the question. Have a nice day.</span>
I think it’s because from earth things may appear smaller smaller in space where in actuality in space it can be way larger
Answer:
1. 43.44g of HCl
2. 26.67 L of HCl
Explanation:
1) Molarity of a solution = number of moles (n) ÷ Volume (V)
According to the provided information in this question,
V = 350 mL = 350/1000 = 0.350L
Molarity = 3.4 M
Using Molarity = n/V
3.4 = n/0.350
n = 3.4 × 0.350
n = 1.19mol
Using the formula below to calculate the mass of HCl;
mole = mass/molar mass
Molar mass of HCl = 1 + 35.5 = 36.5g/mol
mole = mass/MM
mass = 1.19 mol × 36.5g/mol
mass = 43.44g of HCl
2) At STP, HCl has a pressure of 1atm, a temperature of 273K
V = ?
n = 1.19 mol
R = 0.0821 Latm/molK
Using PV = nRT
V = nRT/P
V = 1.19 × 0.0821 × 273/1
Volume = 26.67L
<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.