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3 years ago

How many equivalents are present in 13.5g of Al3+

1 answer:

4 0

Answer:

= 1.5 eq

Explanation:

One definition of an equivalent weight is that it is mass of a substance that gains or loses 1 mole of electrons.
Al3+ has lost 3 e-, so there are 3 equivalent weights in 1 mol Al3+.

1 mol Al3+ =3 eq. wts.
1 mol Al x(27 g / 1 mol)x(1 mol / 3 eq. wts.) = 9.0 g = 1 eq. wts.

13.5 g Al3 + x (1 eq.wt. / 9.0 g) = 1.5 eq

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If molecules move from one place to another inside of a cell, which type of communication is this?

C.intracellular, enzyme-linked receptor.

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3 years ago

29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

arsen [322]

The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

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3 years ago

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Answer:

D. It contains a phosphate with higher phosphoryl transfer potential than ATP

Explanation:

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What is a form of potential energy that is due to relative positions of the particles within a materials.

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I think is D I hope it helped :)

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3 years ago

Predict the splitting pattern for each of the labeled hydrogens in the following molecules. Assume that all coupling constants a

Ghella [55]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

b) Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

c) The IUPAC name is Butan-2-ol

Explanation:

Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.

According to that for molecule 1 as shown on the second uploaded image

For Proton Ha, number of non equivalent protons n= 2

Splitting pattern for Ha= 2+1 , Triplet

For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3

Splitting pattern for Hb and Hc= 3+1 , Quartet

For Proton Hd, number of non equivalent protons n= 0

Splitting pattern for Hd= 0+1 , Singlet

Considering the second question for Molecule 2 as shown on the third uploaded image

For Proton Ha, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

For Proton Hb, number of non equivalent protons n= 3

Splitting pattern for Hb= 3+1=4 , Quartet

For Proton Hc, number of non equivalent protons n= 3

Splitting pattern for Hc= 3+1=4 , Quartet

For Proton Hd, number of non equivalent protons n= 1

Splitting pattern for Ha= 1+1=2 , Doublet

Considering the third question

The name of the given molecule is gotten according to longest carbon chain = 4 (Prefix 'Butan')

Functional group = -OH (Suffix 'ol') at C-2

The IUPAC name is Butan-2-ol

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3 years ago