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IgorC [24]
3 years ago
5

How many moles are there in 1.09 kg of Al(OH)3?

Chemistry
1 answer:
alina1380 [7]3 years ago
4 0
  question   1
moles  =  mass/molar  mass  of  Al(OH)3
convert   Kg  to  g
that  is  1.09  x  1000=1090g

moles  is  therefore=1090g/78(molar  mass  of Al(OH)3)=  13.974 moles

   question  2
moles=
2.55g/327.2(molar   mass  of  Pb(CO3)2=  7.79  x   10^-3   moles
from  avogadro  constant
1moles=6.02  x10^23  formula  units
what  about  7.79  x  10  ^-3
={(7.79  x  10^-3)moles  x ( 6.02  x10^23)} /1  mole=4.69  x10^21  formula  units
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trapecia [35]

Answer:

None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.

Explanation:

Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.

Before balanced Left side.

Cl-2

O-8

H-2

Before balanced right side.

H-1

Cl-1

O-3

That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.  

(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)

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2 years ago
Look at the four positions of Earth with respect to the sun. Florida is in the Northern hemisphere. At what position of Earth wi
Lena [83]

Answer:

2) Position 2 3

Explanation:

3 0
3 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

6 0
3 years ago
_Mg+_O2=_MgO<br> Balance out this reaction above
Ira Lisetskai [31]
2 Mg + 1 O2 = 2 MgO.
8 0
2 years ago
Convert 1.248×1010 g to each of the following units.
Leona [35]

Answer:

a) 1.248 x 10⁷ kg

b) 1.248 x 10⁴ Mg

c) 1.248 x 10¹³ mg

d) 1.248 x 10⁴ ton

Explanation:

a) Since 1000 g = 1 kg we can convert grams to kg by multiplyig any given quantity in grams by the conversion factor ( 1 kg / 1000 g):

1.248 x 10¹⁰ g * (1 kg / 1000 g) = 1.248 x 10⁷ kg

b) Since 1 Mg = 1 x 10⁶ g, the conversion factor will be ( 1 Mg / 1 x 10⁶ g):

1.248 x 10¹⁰ g * (  1 Mg / 1 x 10⁶ g) = 1.248 x 10⁴ Mg

c) Since 1 mg = 1 x 10⁻³ g, the conversion factor will be ( 1 mg / 1 x 10⁻³ g):

1.248 x 10¹⁰ g ( 1 mg / 1 x 10⁻³ g) = 1.248 x 10¹³ mg

d) Since 1 metric ton = 1000 kg and 1000 g = 1 kg, we can use these conversions factors: ( 1 kg / 1000 g) and (1 ton / 1000 kg):

1.248 x 10¹⁰ g * ( 1 kg / 1000 g) * ( 1 ton / 1000 kg) = 1.248 x 10⁴ ton

8 0
2 years ago
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