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Genrish500 [490]
3 years ago
12

How many ammonium ions, nh4 , are there in 5.0 mol (nh4)2s?

Chemistry
2 answers:
aleksandr82 [10.1K]3 years ago
5 0

<u>Answer:</u> The number of ammonium ions in given moles of ammonium sulfide are 6.022\times 10^{24}  

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of (NH_4)_2S

In 1 mole of ammonium sulfide, 2 moles of ammonium ions and 1 mole of sulfide ions are present.

According to mole concept:  

1 mole of compound contains 6.022\times 10^{23} number of atoms

So, 5 moles of ammonium sulfide molecule will contain = 5\times 2\times 6.022\times 10^{23}=6.022\times 10^{24} number of ammonium ions.

Hence, the number of ammonium ions in given moles of ammonium sulfide are 6.022\times 10^{24}

olchik [2.2K]3 years ago
3 0
There are two such ions in every molecule
1 mol contains 6.02 * 10^23 things.
1 mol of (NH4)2S  contains 2*6.02 * 10^23 ions (of NH4+ in this case.) = 1.204 * 10^24 ions HN4+

5 mol of (NH4)2S contains 5 * 1.204*10^24 =
6.02 * 10^24 mols NH4+ions <<<<====Answer
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Which of the following statements is a true statement regarding a solution with [H1+] =1x10-5 M and [OH1-]= 1x10-9 M?
zheka24 [161]

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B. The [H1+] >[OH1-] and the solution is acidic

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2 years ago
What is the mole ratio of D to A in the generic chemical reaction 3A + B C + 4D
ehidna [41]

Answer:- Mole ratio of D to A is 4:3.

Explanations:- Mole ratio for a chemical reaction is the ratio of the coefficients.

The given generic chemical reaction is:

3A+B\rightarrow C+4D

The numbers written in front of each chemical species in the chemical reaction are their moles. For the given generic chemical reaction the coefficient of A is 3 and that of B is 1. So, the mole ratio of A to B is 3:1.

Similarly if we want to write the mole ratio of C to D then it is 1:4.

We are asked to write the mole ratio of D to A. So, like the other ratios, the mole ratio of D to A is 4:3 as the coefficient of D is 4 and A is 3.

7 0
3 years ago
Describe briefly how to obtain the radial probability of an electron​
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Explanation:

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8 0
2 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

                                 1 mol of CoSO4 -------------- 2 moles of NaOH

                                0.092 moles      ---------------   x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

6 0
3 years ago
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