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CaHeK987 [17]
3 years ago
8

How many hydrogen atoms are present in a hydrocarbon chain of five carbon atoms with two double bonds and two single bonds?

Chemistry
1 answer:
VashaNatasha [74]3 years ago
7 0
Is it multiple choice? if not, five.

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Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the ch
geniusboy [140]
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2
3 0
3 years ago
How many protons z and how many neutrons n are there in a nucleus of the most common isotope of silicon, 2814si?
zheka24 [161]
The 28 is the sum of the protons and neutrons in the element silicon.
ALL silicon atoms have 14 protons in the nucleus, so we can turn this into an equation:
#protons + #neutrons = 28
14 + #neutrons = 28
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7 0
3 years ago
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Which list of elements is arranged in order of increasing electronegativity?
muminat
K, ca, sc is the right answer. Take a look at table S of your chemistry reference table. 
6 0
3 years ago
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Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
Which of the following steps will speed up the reaction between a baking soda solution and a calcium chloride solution in a beak
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I think keeping it in the sunlight.
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