initial volume of the argon sample = 5.93L according to Boyle's law
What is Boyle's law ?
Boyle's law, also known as Mariotte's law, is a relationship describing how a gas will compress and expand at a constant temperature. The pressure (p) of a given quantity of gas changes inversely with its volume (v) at constant temperature, according to this empirical connection, which was established by the physicist Robert Boyle in 1662. In equation form, this means that pv = k, a constant.
According to Boyle's law
P1/V1 = P2/V2
P1 = initial pressure
P2 = final pressure
V1 =initial volume
V2= final volume
V1 = P1*V2/P2
V1 = 2.32*18.3/7.16 = 5.93L
initial volume of the argon sample = 5.93L according to Boyle's law
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Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu
Answer:
Cl2 bonds are stronger!!!!!
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
