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Alex Ar [27]
2 years ago
9

Will mark a brainliest.

Chemistry
2 answers:
olga2289 [7]2 years ago
6 0

Answer:

C2O4

Explanation:

Di means 2 and tetra means 4

ch4aika [34]2 years ago
5 0

Answer:

The answer is actually CO4

Explanation:

prove me wrong

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how many moles of sodium are needed to react with sulfuric acid to produce 3.75 moles of sodium sulfate according to the followi
anastassius [24]
Hello!
<span>
You'll need to react 7,5 moles of Sodium with sulfuric acid to produce 3.75 moles of sodium sulfate
</span>
First of all, you need to balance the reaction. The balanced reaction is shown below (ensuring that the Law of Conservation of Mass is met on both sides):

2Na + H₂SO₄ → Na₂SO₄ + H₂

Now, all that you have to do is to use molar equivalences in this reaction applying the coefficients to calculate the moles of Sodium that you'll need:

molesNa=3,75moles Na_{2} SO_4* \frac{2 moles Na}{1 mol Na_{2} SO_4} =7,5 moles Na 

Have a nice day!
5 0
3 years ago
If an ultraviolet photon has a wavelength of 77.8 nm calculate the energy of one mole ultraviolet photon.
DerKrebs [107]

Answer:

Explanation:

E = (hc)/(λ)

E = (6.624x10^(-27))Js x ((3×10^8)ms^(-1)) /

(77.8x10^(-9)m)

E = 2.55 x 10^(-11) J

7 0
2 years ago
Please help on 8 thank youuuuuuuuuu
enyata [817]

Answer:

b)5l x 10kg  c)10kg + 9l   (Not sure for the last 1)

6 0
3 years ago
Which element has chemical properties that are most<br> similar to potassium, And why?
bekas [8.4K]

Answer:

Brainliest pls

Explanation:

The components potassium and sodium have comparable substance properties since they have a similar number of valence electrons

8 0
2 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
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