How many atoms are in 0.3500 mol of gold

Chemistry

1 answer:

Vikentia [17]3 years ago

3 0

8798076 atoms are in a 0.3500 mol of gold

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1093.5 rounded into 3 sig figs

Ray Of Light [21]

Significant figures means the number of important figures, 0 not included.

This means that the answer is 1094.00

Hope this helps! :)

3 0

3 years ago

Solid mixture contains MgCl2 and NaCl. When 0.5000 g of this solid is dissolved in enough water to form 1.000 L of solution, the

aksik [14]

Answer:

83,40 (w/w) %

Explanation:

The osmotic pressure (π) is defined as:

π = iMRT

Where i is Van't Hoff factor (3 for MgCl₂ and 2 for NaCl), M is molarity, R is gas contant (0,082atmL/molK) and T is temperature (298,15K)

iM = π / RT

iM = 0,01607mol/L

It is possible to write:

3x+2y = 0,01607mol/L (1)

Where x are moles of MgCl₂ and y moles of NaCl.

-M = moles of each compund because M is molarity (moles/L) and there is 1,000L-

Knowing molar mass of MgCl₂ is 95,2 g/mol and for NaCl is 58,44g/mol:

x×95,211 + y×58,44 = 0,5000g (2)

Replacing (2) in (1):

3x+2(0,0086 - 1,629x) = 0,01607mol/L

-0,258x = -0,00113

x = 0,004380 moles of MgCl₂

In grams:

0,004380 moles of MgCl₂×(95,211g/mol) = 0,417g of MgCl₂

Mass percent is:

(0,4170g of MgCl₂/0,5000g of solid) ×100 = 83,40 (w/w) %

I hope it helps!

4 0

3 years ago

Find the equilibrium value of [CO] if Kc=14.5 : CO (g) + 2H2 (g) ↔ CH3OH (g) Equilibrium concentrations: [H2] = 0.322 M and [CH3

Annette [7]

Answer:

The equilibrium value of [CO] is 1.04 M

Explanation:

Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.

Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:

aA + bB ⇔ cC + dD

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case:

$Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }$