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2 years ago

You have a sample containing 2.39 moles of silver. How many atoms of silver are in this sample?

Chemistry

1 answer:

soldi70 [24.7K]2 years ago

4 0

Answer:

14.3878 x 10^23 molecules Ag

Explanation:

2.39 mol Ag 6.02 x 10^23

----------------- x ------------------ = 14.3878 x 10^23 Molecules Ag

1 1 mol

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sp2606 [1]

Inbox me for the answer

6 0

3 years ago

The same reaction is begun with an initial concentration of 0.05 M O3 and 0.02 M NO. Under these conditions, the reaction reache

Ivahew [28]

The balanced equation of the reaction is:

O3(g) + NO (g) → O2 (g) + NO2 (g)

Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2

If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.

The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.

And the rate of reaction is change in concetration divided by the time.

The change in concentration in O3 is 0.02 M

Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s

8 0

3 years ago

Can someone please explain the relationship between the group number, the number of valence electrons lost or gained, and the fo

soldi70 [24.7K]

<span>Ca has 2 valence electrons and F has 7, they would end up having to share electrons but Ca or F has to keep one because in total there are 9 valence elections and only 8 are needed to make this pair "happy". that's all i can provide. i hope it helped some</span>

6 0

3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as

zubka84 [21]

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g 22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

= 175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

6 0

3 years 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

3 years ago

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