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Shtirlitz [24]
3 years ago
9

What are the missing coefficients for the skeleton equation below? cr(s) + fe(no3)2(aq) → fe(s) + cr(no3)3(aq)?

Chemistry
1 answer:
strojnjashka [21]3 years ago
8 0
SThe  missing   coefficient  for  the  skeleton   equation  below  is  as  follows

skeleton   equation

Cr(s)  +  Fe(No3)2(aq)  ------> Fe (s)   +  Cr(NO3)3  (aq)
the  missing  coefficient  are  is   as  follows

 2 Cr(s)   +  3  Fe(NO3)2  ---> 3 Fe (s)  +  2 Cr(NO3)3

This  is  obtained   by  making  sure  all  the   molecules  are  balanced  in  both  sides
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Describe the properties of alkali metals. based on their electronic arrangement, explain whether they exist alone in nature.
saul85 [17]

The alkali metals can't exist alone in nature because of incomplete outermost shell of alkali metals.

<h3>What are the  properties of alkali metals?</h3>

The alkali metals have the high thermal and electrical conductivity. It has high lustre, ductility, and malleability as compared to other materials. Each alkali metal atom has one electron in its outermost shell which make more reactive.

So we can conclude that the alkali metals can't exist alone in nature because of incomplete outermost shell of alkali metals.

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5 0
2 years ago
. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:
mafiozo [28]

Answer:

See explanation

Explanation:

a) The equation of the reaction is;

2Na + Cl2 ------>2NaCl

Number of moles of sodium = 10g/23 g/mol = 0.43 moles

If 2 moles of sodium reacts with 1 mole of Cl2

0.43 moles reacts with 0.43 * 1/2 = 0.215 moles of Cl2

Mass =  0.215 moles of Cl2 *71 g/mol = 15.265 g

b) Equation of the reaction;

HgO -> Hg + O2

1.252 moles of HgO 1.252/32 gmol = 0.039 moles

1 mole of HgO  yields 1 mole of oxygen hence

0.039 moles of HgO yields   0.039 moles of oxygen

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c) Equation of the reaction;

2NaNO3 -----> 2NaNO2 + O2

Number of moles of 128 g of oxygen = 128g/32 g/mol = 4 moles

2 moles of NaNO3  yields 1 mole of oxygen

x moles of NaNO3   yields 4 moles of oxygen

x = 8 moles of NaNo3

Mass of NaNO3 = 8 * 85 g/mol = 680 g of NaNo3

6 0
3 years ago
A sample of phosphonitrilic bromide, PNBr2, contains 2.01 mol of the compound. Determine the amount (in mol) of each element pre
nordsb [41]

Answer:

2.01 moles of P → 1.21×10²⁴ atoms

2.01 moles of N → 1.21×10²⁴ atoms

4.02 moles of Br → 2.42×10²⁴ atoms

Explanation:

We begin from this relation:

1 mol of PNBr₂ has 1 mol of P, 1 mol of N and 2 moles of Br

Then 2.01 moles of PNBr₂ will have:

2.01 moles of P

2.01 moles of N

4.02 moles of Br

To determine the number of atoms, we use the relation:

1 mol has NA (6.02×10²³) atoms

Then: 2.01 moles of P will have (2.01  . NA) = 1.21×10²⁴ atoms

2.01 moles of N (2.01  . NA) = 1.21×10²⁴ atoms

4.02 moles of Br (4.02 . NA) = 2.42×10²⁴ atoms

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3 years ago
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