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2 years ago

Give the name and formula of the compound formed from each pair of elements: (a) 37q and 35r (b) 8q and 13r (c) 20q and 53r

2 answers:

amm18122 years ago

8 0

(a) 37Q = 37Rb(Rubidium)
35R = 35Br(Bromine)
Formula of the compound is : RbBr(Rubidium Bromide)

(b) 8Q = 8O(Oxygen)
13R = 13Al(Aluminum)
Formula of the compound is : Al2O3(Aluminum Oxide)

(c)20Q = 20Ca(Calcium)
53R = 53I(Iodine)
Formula of the compound is : CaI2(Calcium Iodide)

HACTEHA [7]2 years ago

6 0

Answer:

(a) $RbBr$ - rubidium bromide.

(b) $Al_2O_3$ - aluminium oxide.

Explanation:

Hello,

(a) In this case, the elements having 37 and 35 as the oxidation states are rubidium and bromine respectively, in such a way the formed compound, due their gatherable oxidation states (+1 and -1 respectively) is the following binary salt:

$RbBr$ - rubidium bromide.

(b) In this case, the elements having 8 and 13 as the oxidation states are aluminium and oxygen respectively, in such a way the formed compound, due their gatherable oxidation states (+3 and -2 respectively) is the following basic oxide:

$Al_2O_3$ - aluminium oxide.

(c) In this case, the elements having 20 and 53 as the oxidation states are calcium and iodine respectively, in such a way the formed compound, due their gatherable oxidation states (+2 and -1 respectively) is the following binary salt:

$CaI_2$ - Calcium iodide.

Best regards.

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What can you summarize about the use of acid-base indicators?

densk [106]

An acid-base indicator is used to identify the presence of an acid or base. These indicators exhibit different colors depending on the solution that they are in. They are especially useful when completing titrations to determine the molarity of an unknown substance and is denoted as option D.

<h3>What is Titration?</h3>

This is done in the laboratory and involves the slow addition of one solution of a known concentration to a known volume of another solution of unknown concentration.

Acid-base indicators are used to determine the presence of an acid or base in a solution which is based on the colors seen when performing the chemical reaction.

It is used to calculate the the molarity of an unknown substance through the knowledge of the other parameters which is therefore the reason why option D was chosen as the most appropriate choice.

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7 months ago

Rotation about a carbon-carbon double bond does not readily occur because: __________.1) the overlap of the p orbitals of the ca

Advocard [28]

Answer:

1) The overlap of the p orbitals of the carbon-carbon π bond would be lost

Explanation:

Unlike simple bonds, a double bond can not rotate, since it is not possible to twist the ends of the molecule without breaking the π bond.

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2 years ago

Which molecule transports oxygen in red blood cells? (1 point)

Afina-wow [57]

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3 years ago

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 $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

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2 years ago

What property do atoms of these elements have that helps make the molecules polar

DENIUS [597]

You have not mention here about those elements but the general concept for this is the:
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even distribution -------> non-polar

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2 years ago