<span>Answer: 17.8 cm
</span>
<span>Explanation:
</span>
<span>1) Since temperature is constant, you use Boyle's law:
</span>
<span>PV = constant => P₁V₁ = P₂V₂
</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:
</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>
<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:
</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³
</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>
1,38×10²² = 0,138×10²³
0,138×10²³ ----- 1,5g
6,02×10²³ ------ X
X = (1.5×6,02×10²³)/0,138×10²³
X = 65,435 g/mol
It's ZINC (Zn)
:•)
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
The molar mass of M is 0.225g/mol and the element M is Hydrogen
If a metal M combines with an oxygen element to form the oxide, 
then the chemical reaction will be expressed as:
This shows that 4 moles of an unknown element M react with the oxygen element to produce the oxide
Given the following parameters
Mass of M = 0.890 grams
Mass of = 0.956 grams
Get the molar mass of M:
Molar mass = Mass/number of moles
Molar mass = 0.890/4
Molar mass = 0.225g/mol
Hence the molar mass of M is 0.225g/mol and the element M is Hydrogen
Learn more here: brainly.com/question/6996520
Answer:
B) 0.32 %
Explanation:
Given that:
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)
Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
