All categories

3 years ago

How many atoms are in 12.9 mol Au?

Chemistry

2 answers:

julia-pushkina [17]3 years ago

8 0

Answer:

amy what happen to your old account and the answer is c

Explanation:

REY [17]3 years ago

7 0

Given :

12.9 mol of Au.

To Find :

How many atoms are in 12.9 moles Au.

Solution :

We know, 1 mol of any element/molecules contains $6.022\times 10^{23}\ atoms/mol$ .

Number of atoms in 12.9 moles of Au is :

$n = 12.9 \times 6.022\times 10^{23}\ atoms\\\\n = 7.77\times 10^{24}\ atoms$

Therefore, number of atoms in 12.9 moles of Au is $7.77\times 10^{24}\ atoms$ .

Hence, this is the required solution.

You might be interested in

How many carbon atoms are present in a mol of 12C?

Andrei [34K]

One mole of any substance contains 6.02 x 10²³ particles

5 0

4 years ago

At 293 K, methanol has a vapor pressure of 97.7 Torr and ethanol has a vapor pressure of 44.6 Torr. What would be the vapor pres

Bad White [126]

Answer: The vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction in solution

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$ $p_{total}=x_Ap_A^0+x_BP_B^0$

moles of ethanol= $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{80g}{46g/mol}=1.7moles$

moles of methanol= $\frac{\text{Given mass}}{\text {Molar mass}}=\frac{97g}{32g/mol}=3.0moles$

Total moles = moles of ethanol + moles of methanol = 1.7 +3.0 = 4.7

$x_{ethanol}=\frac{1.7}{4.7}=0.36,$

$x_{methanol}=1-x_{ethanol}=1-0.36=0.64$

$p_{ethanol}^0=44.6torr$

$p_{methanol}^0=97.7torr$

$p_{total}=0.36\times 44.6+0.64\times 97.7=78.3torr$

Thus the vapor pressure of a mixture of 80 g of ethanol and 97 g of methanol at 293 K is 78.3 torr.

8 0

4 years ago

What’s the answer to number 34?

Galina-37 [17]

c- it says I need to write at least 20 words to submit it so- here haha

3 0

4 years ago

. The net ionic equation for the reaction that occurs during the titration of nitrous acid with sodium hydroxide is (a) HNO2 Na

hjlf

Answer:

$H^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}$

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Only the species which are present in aqueous state dissociate.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a)

The balanced molecular equation will be,

$HNO_2_{(aq)}+NaOH_{(aq)}\rightarrow H_2O_{(l)}+NaNO_2_{(aq)}$

The complete ionic equation in separated aqueous solution will be,

$H^+_{(aq)}+NO_2^-_{(aq)}+Na^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}+Na^+_{(aq)}+NO_2^-_{(aq)}$

In this equation the species present are, $Na^+\text{ and }NO_2^-$ are the spectator ions.

Hence, the net ionic equation contains specie is

$H^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}$

6 0

3 years ago

if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius

Lady_Fox [76]

The mass of ore required is 21 700 t.

r = 750 cm

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3} \pi (750 cm)^{3}$ = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ × $\frac{11.34 g}{1 cm^{3} }$ = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g × $\frac{1 kg}{1000 g}$ = 2.004 × 10⁷ kg

2.004 × 10⁷ kg × $\frac{1 t}{1000 kg}$ = 2.004 × 10⁴ t

92.5% efficiency means 92.5 t Pb per 100 t of ore.

Mass of ore = 2.004 × 10⁴ t Pb × $\frac{100 tore}{92.5 t Pb}$ = 2.17 × 10⁴ t ore = 21 700 t ore

6 0

3 years ago