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3 years ago

How to get proton? I need help, its due today

Chemistry

2 answers:

Nana76 [90]3 years ago

8 0

Answer:

The atomic number = number of proton

Nickel = 28 protons

Explanation:

DaniilM [7]3 years ago

8 0

so the protons of nickel is 28

but if you want to find the proton, atomic number is equal to proton

atomic number = proton

hope it helps !

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What substance is acting as the Brønsted-Lowry base in the forward reaction below? H2O + HCI ----> H3O+ CI-

yarga [219]

H2O is the Bronsted-Lowry base because it accepts the hydrogen ion to become H3O after the reaction is complete.

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3 years ago

DENIUS [597]

Answer:

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Explanation:

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3 years ago

Lithium ions in Lithium selenide (Li2Se) have an atomic radius of 73 pm whereas the selenium ion is 184 pm. This compound is mos

NARA [144]

Explanation:

Formula according to the radius ratio rule is as follows.

$\frac{r_{+}}{r_{-}} = \frac{73}{184}$

= 0.397

According to the radius ratio rule, as the calculated value is 0.397 and it lies in between 0.225 to 0.414. Therefore, it means that the type of void is tetrahedral.

Thus, we can conclude that the given compound is most likely to adopt closest-packed array with lithium ions occupying tetrahedral holes.

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3 years ago

Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi

AlexFokin [52]

Explanation:

The given cell reaction is as follows.

$In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)$

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : $In(s) \rightarrow In^{3+}(aq) + 3e^{-}$ ...... (1)

At cathode; Reduction-half reaction : $Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s)$ ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

$2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}$

$3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)$

Net reaction: $2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

Thus, we can conclude that the overall cell reaction is as follows.

$2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)$

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3 years ago

How many grams of product can be produced by reacting 5.0 gram of aluminum and 22 grams of bromine? 2 Al + 3 Br2 > 2 AlBr3

Crazy boy [7]

Answer:

The mass of AlBr3 is 24.5 grams

Explanation:

Step 1: Data given

Mass of aluminium = 5.0 grams

Mass of bromine = 22.0 grams

Molar mass of aluminium = 26.98 g/mol

Molar mass of br2= 159.8 g/mol

Step 2: The balancced equation

2 Al + 3 Br2 → 2 AlBr3

Step 3: Calculate moles Al

Moles Al = mass Al / molar mass Al

Moles Al = 5.0 grams / 26.98 g/mol

Moles Al = 0.185 moles

Step 4: Calculate moles Br

Moles Br = 22.0 grams / 159.8 g/mol

Moles Br = 0.138 moles

Step 5: Calculate limiting reactant

For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3

Br2 is the limiting reactant. It will completely be consumed (0.0313 moles)

Al is in excess. There will react 2/3*0.138 = 0.092 moles

There will remain 0.185- 0.092 = 0.093 moles Al

Step 6: Calculate moles AlBr3

For 2 moles Al we need 3 moles Br2 to produce 2 moles AlBr3

For 0.0313 moles Br2 we'll have 2/3*0.138 = 0.092 moles AlBr3

Step 7: Calculate mass AlBr3

Mass AlBr3 = moles * molar mass

Mass AlBr3 = 0.092 moles * 266.69 g/mol

Mass AlBr3 = 24.5 grams

The mass of AlBr3 is 24.5 grams

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3 years ago