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mrs_skeptik [129]
3 years ago
9

How many valence electrons does Arsenic have?

Chemistry
1 answer:
adoni [48]3 years ago
4 0
Arsenic's atomic number is 33.

Hence his electronic configuration is 1s^22s^22p^63s^23p hence his valence electrons are 4s^24p^3.

5 valence electrons
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PLEASE HELP, I NEED TO GET THIS DONE TODAY
luda_lava [24]

The balanced reaction is

3Na3PO4  +  2CuSO4  ------>  3Na2SO4  + Cu3(PO4)2

To balance this reaction of double displacement, we see first that this reaction maintain the valence numbers of every atom.

Then, to have the same value of Na in the two sides of the reaction we multiply for the number of the other side. So,

(Na3PO4)x 2

(Na2SO4)x3

As we can see either, we need to balance PO4 cause there are two molecules of this in the reactant side, so we have two molecules of PO4 in the product either.

Then we get

3Na3PO4  +  2CuSO4  ------>  3Na2SO4  + Cu3(PO4)2

To probe that balance was correct, you can verify that the charges are exactly the opposite.

4 0
3 years ago
Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta
AysviL [449]

<u>Answer:</u> The \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]

We are given:

\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ

Putting values in above equation, we get:

-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ

Hence, the \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

8 0
3 years ago
. In lab we used 40 ml of 6M NaOH in making soap.<br>How many moles of NaOH was in this volume?​
Firdavs [7]

Hey there!:

Molarity of NaOH = 6 M or 6 mol/L

Volume of NaOH = 40 mL

Therefore , number of moles of NaOH:

40 mL  =( 6 mol / 1000 mL)*40=

6/ 1000 * 40 =>  0.24 moles of NaOH

Hope this helps!

8 0
3 years ago
How is dinitrogen (1) oxide distinguished shed from oxygen​
grigory [225]

Answer:

See Explanation

Explanation:

Both oxygen and dinitrogen (1) oxide are known to rekindle a glowing splint. However, oxygen is an odorless gas while dinitrogen (1) oxide has a faint smell and has an effect on the central nervous system.

Also, the combustion of dinitrogen (1) oxide produces brown fumes of nitrogen IV oxide.

Finally, when dinitrogen (1) oxide and oxgen are both heated with copper, residual nitrogen gas is left with dinitrogen (1) oxide while no residual gas is left with oxygen.

4 0
3 years ago
Convert 35 joules into calories.<br> OR 1.410 cal<br> OC 82,500 cal<br> OD. 1.420,000 cal
lesya692 [45]

Answer:

0.83Cal

Explanation:

Given parameters:

   Amount of energy given;

      35J;

 Convert to Cal;

         1 calorie  = 42 joules

         x calories  = 35 joules

   Cross multiply;

         42x = 35

             x = \frac{35}{42}   = 0.83Cal

7 0
3 years ago
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