Rubidium is the alkali metal that would melt on a hot day.
Answer:
12 moles of water will be produced
Explanation:
Given data:
Number of moles of NH₃ = 8.00 mol
Number of moles of O₂ = 14.0 mol
Number of moles of H₂O produced = ?
Solution:
Chemical equation:
4NH₃ + 7O₂ → 4NO₂ +6H₂O
Now we will compare the moles of reactant with product.
NH₃ : H₂O
4 : 6
8 : 6/4×8 = 12
O₂ : H₂O
7 : 6
14 : 6/7×14 = 12
12 moles of water will be produced.
Answer:
ΔT = 76.5 °C
Explanation:
Given data:
Amount of water = 100.0 g
Energy needed = 32000 J
Change in temperature = ?
Solution,
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values in formula.
Q = m.c. ΔT
ΔT = Q / m.c
ΔT = 32000 j/ 100.0 g × 4.184 j/g. °C
ΔT = 32000 j / 418.4 j /°C
ΔT = 76.5 °C
<h3>Answer:</h3>
Volume = 11.2 L
<h3>Explanation:</h3>
Step 1: Calculate Moles:
As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.
The relation between Moles, Number of Particles and Avogadro's Number is given as,
Number of Moles = Number of Particles ÷ 6.022 × 10²³
Putting values,
Number of Moles = 3.01× 10²³ Particles ÷ 6.022 × 10²³
Number of Moles = 0.50 Moles
Step 2: Calculate Volume:
As we know that one mole of any Ideal gas at standard temperature and pressure occupies exactly 22.4 dm³ volume.
When 1 mole gas occupies 22.4 dm³ at STP then the volume occupied by 0.50 moles of gas is calculated as,
= (22.4 dm³ × 0.50 moles) ÷ 1 mole
= 11.2 dm³ ∴ 1dm³ = 1 L
So,
Volume = 11.2 L
Answer:
The abundance of first isotope is 69.15 %
The abundance of second isotope is 30.85 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope:
% = x %
Mass = 62.9296 u
For second isotope:
% = 100 - x
Mass = 64.9278 u
Given, Average Mass = 63.546 u
Thus,
Solving for x, we get that:
x = 69.15 %
<u>The abundance of first isotope is 69.15 %</u>
<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>
