Which is the force of attraction that holds atoms together?

If 4.50 l of water vapor at 50.2 °c and 0.121 atm reacts with excess iron, how many grams of iron(iii) oxide will be produced?

Flura [38]

When the balanced equation for this reaction is:
2Fe + 3H2O → Fe2O3 + 3H2

and according to the vapour pressure formula:
PV= nRT
when we have P is the vapor pressure of H2O= 0.121 atm
and V is the volume of H2O = 4.5 L
and T in Kelvin = 52.5 +273 = 325.5 K
R= 0.08205 atm-L/g mol-K
So we can get n H2O
So, by substitution:
n H2O = PV/RT
= (0.121*4.5)/(0.08205 * 325.5) = 0.02038 gmol
n Fe2O3 = 0.02038 * (1Fe2O3/ 3H2O) = 0.00679 gmol
Note: we get (1FeO3/3H2O) ratio from the balanced equation.
we can get the Mass of Fe2O3 from this formula:
Mass = number of moles * molecular weight
when we have a molecular weight of Fe2O3 = 159.7
= 0.00679 * 159.7 = 1.084 g
∴ 1.084 gm of Fe2O3 will produced

6 0

4 years ago

Read 2 more answers

Calculate the mass percent of oxygen in CO2 (Work shown)

Genrish500 [490]

Answer:

$mass \: of \: CO _{2} = (12 + (16 \times 2)) = 44 \: g \\ mass \: of \: oxygen \: in \: CO _{2} = (16 \times 2) = 32 \\ \% \: of \: oxygen = \frac{32}{44} \times 100\% \\ = 72.7\%$

4 0

3 years ago

Chem help 36 and 37 the word cut off on 37 is larger

ad-work [718]

The solution for problem 36 is in the picture attached. I am 80% sure it’s correct.

7 0

3 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

olasank [31]

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

_________________________________
_________________________________

Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

7 0

3 years ago

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol

defon

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $\frac{[A^{-}]}{[HA]}$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $\frac{[A^{-}]}{[HA]}$

1,38 = $\frac{[A^{-}]}{[HA]}$ (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $\frac{60,05 g}{1mol}$ = 9,0 g of acetic acid

I hope it helps!

5 0

3 years ago