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3 years ago

Please Help!! Qns 1b to 3

Chemistry

1 answer:

Gelneren [198K]3 years ago

7 0

Answer: Ammonium, when heated with aqueous base, will give off NH3 (ammonia) gas, (and depending, water vapor). This will leave the Cr2O3(s). From then on,

it is just adding or subtraction of gases or water vapor. You probably heard “Loss of electrons is Oxidation”, “Gain of Electrons is reduction”. That should help.

Explanation: This isn’t an explanation but an interesting point; Acid-Base and RedOx reactions are useful to the most complex of any Chemistry. Get this down, and Organic Chemistry will be much easier.

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Balance the following skeleton ionic reaction, and calculate E°cell to decide whether the reaction is spontaneous: Ag(s) + Cu2+

tia_tia [17]

Answer:

2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)

Explanation:

Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.

E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V

The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.

5 0

3 years ago

Read 2 more answers

What geological process(es) could form magnesium silicate perovskite at 500 C?

dimulka [17.4K]

Magnesium silicate is a compound of magnesium oxide and silicon. It is the magnesium salt of silicic acid containing an unspecified amount of water. The molecular formula can be expressed more clearly as MgSiO3.

7 0

1 year ago

Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-

hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written between $Zn^{2+} (Positive ion)and F^{-} (Negative ion)$

First Formula would be $ZnF_{2}$

Second empirical formula is between $Zn^{2+}(Positive ion) and O^{2-}(Negative ion)$

Second Formula would be $Zn_{2}O_{2}$

Note : When the subscript are same they get cancel out, so $Zn_{2}O_{2}$ would be written as $ZnO$

Third empirical formula is between $Ni^{4+}(Positive ion) and F^{-}(Negative ion)$

Third Formula would be : $NiF_{4}$

Forth empirical formula is between $Ni^{4+}(Positive ion)and O^{2-}(negative ion)$

Forth Formula would be : $Ni_{2}O_{4}$ or $NiO_{2}$

Note- The subscript will be simplified and the formula will be written as $NiO_{2}$ .

The empirical formula of four binary ionic compounds are : $ZnF_{2}, ZnO, NiF_{4},NiO_{2}$

8 0

2 years ago

Read 2 more answers

Identify the products in the equation below (the “I” is Iodine—to differentiate capital “I” from lowercase “l”)

Inessa [10]

Answer : The products are Silver sulfide, $(Ag_2S)$ and Sodium iodide, $(NaI)$ .

Explanation :

The given balanced chemical reaction is,

$2AgI+Na_2S\rightarrow Ag_2S+2NaI$

From the given balanced reaction, we conclude that the 2 moles of silver iodide react with the 1 mole of sodium sulfide to give product as 1 mole of silver sulfide and 2 moles of sodium iodide.

In a chemical reaction, reactants are represent on the left side of the right-arrow and products are represent on the right side of the right-arrow.

Therefore, in a chemical reaction the products are Silver sulfide and Sodium iodide.

5 0

3 years ago

Chemistry help. Will give Brainliest!

VashaNatasha [74]

#1. An element or ion that has lost two electrons must have a net charge of 2+, because it has two more protons than electrons, therefore the answer is Mg2+

#2. aluminum ions have an oxidation state of 3+ and fluoride has an oxidation state of 1-, therefore I’m order for the charges to cancel you need 3 fluoride ions.
Therefore, the answer is AlF3

8 0

2 years ago