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Daniel [21]
3 years ago
7

___Al+NaOH=>__Na3AlO3+H2

Chemistry
1 answer:
MaRussiya [10]3 years ago
8 0

Explanation:

<em><u>2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2</u></em>

<em><u>...</u></em>

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Where, approximately, is the negative pole on each of these molecules?
Leni [432]

Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.

Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.

5 0
3 years ago
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What is the mass in grams of 2.00 10 5 atoms of naturally occurring neon
lesantik [10]

Answer:

2.00X10^5 x 20gNe/6.02x10^23=6.46x10^-18 but books answer is 797.

Explanation:

3 0
2 years ago
The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

rate constant k =  1.94\times 10^{-4} min^{-1}

Half life t_{1/2} = \frac{0.693}{k} = 3572 min

Initial pressure N_2 O = 4.70 atm

Pressure after 3572 min = P

According to first order kinematics

k = \frac{1}{t} ln\frac{4.70}{P}

1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}

solving for P we get

P = 2.35 atm

2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
Read 2 more answers
Which isomer of 1-bromo-4-tert-butylcyclohexane reacts faster when refluxed with potassium tert-butoxide?
grigory [225]

Answer:

See explanation.

Explanation:

Since potassium tert-butoxide is a strong bulky base, we expect that an elimination reaction predominates.

This reaction is expected to occur by E2 mechanism. The cis isomer of  1-bromo-4-tert-butylcyclohexane is known to react faster by E2 mechanism because it fulfills the anticoplanar arrangement required for for E2 mechanism, where the leaving group and adjacent proton must be anti to each other and in the same plane. The trans isomer can not fulfill this requirement.

7 0
3 years ago
When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
Feliz [49]

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

8 0
3 years ago
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