Answer:
<h2>1.505 × 10²⁴ particles</h2>
Explanation:
The number of particles in iron (II) chloride can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 2.5 × 6.02 × 10²³
We have the final answer as
<h3>1.505 × 10²⁴ particles</h3>
Hope this helps you
Find the number of moles of sodium you have:
<span>n = m/M where m is your 20g of sodium and M is 22.99 g/mol. </span>
<span>Look at the stoichiometry of the equation - it's 2:2 when you are producing NaOH. So if you took 1 mole of Na, it'd produce 1 mole of NaOH (as the ratio is equal). </span>
<span>That means that your moles of sodium is equal to the moles of NaOH produced. Use the molar mass of NaOH - which is 39.998 g/mol along with your calculated number of moles to get the mass (the formula rearranges to m = nM). </span>
<span>This figure is the theoretical yield - what you would get if every last mole of sodium was converted into NaOH. </span>
<span>What you get in practice is the experminetal yield, and the percentage yield is the experimental yield divided by the theoretical yield - and then multiplied by 100%.</span>
Answer:
Volumen = 65.199 metro cúbico
Explanation:
Tal como lo conocemos
P1V1 = P2 V2
Aquí
P1 = 1,9 atmósferas
V1 = (pi) * r ^ 2 * h
Sustituyendo los valores dados, obtenemos
V1 = 3.14 * 2.8 * 2.8 * 4.6 = 113.241 metros cúbicos
V2 = por determinar
P2 = 3.3 atmósferas
Sustituyendo los valores dados y derivados en la fórmula principal, obtenemos -
1.9 atm * 113.241 metros cúbicos = 3.3 atm * V2
V2 = 1.9 atm * 113.241 metros cúbicos / 3.3 atm
V2 = 65.199 metro cúbico
pv=nrt
(.80atm)(?)=(.75)( .0821atm)(50+273)
