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4 years ago

8

How many hydrogen atoms are attached to each carbon adjacent to a double bond? nurition?

1 answer:

larisa [96]4 years ago

3 0

That depends. there are 2 possible answers.
H
C - C = C - H gives a different answer on the right than on the left.

One the left side, the second Carbon is attached to a double bond and has but one hydrogen attached to it.

The Carbon on the right of the double bond has 2
H
C- C = C - H
H

I'm not sure what you should put. It's one of those things that I would repeat my argument and submit it.

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What can you conclude about the relative concentrations of ch3nh2 and ch3nh3cl in this buffer? for ch3nh2, pkb = 3.36?

Alex17521 [72]

POH = pKb + log [salt] / [base] ---> Basic Buffer
and pH = 14 - pOH

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3 years ago

CADMIUM SOLUBILITY AND pH Cadmium is a toxic metal. It can be removed from water by chemical precipitation of solid cadmium hydr

Serga [27]

Explanation:

It is known that,

Molar mass of Cd = 112.41 g/mol

Standard concentration of Cd = 0.005 mg/L = $0.005 \times 10^{-3}$ g/L

Hence, we will calculate the molarity as follows.

Molarity = $\frac{0.005 \times 10^{-3}}{112.41} mol/L$

= $4.45 \times 10^{-8}$ M

Equation for the reaction is as follows.

$Cd(OH)_{2} \rightleftharpoons Cd^{2+} + 2OH^{-}$

$K_{sp} = [Cd^{2+}][OH^{-}]^{2}]$

$2 \times 10^{-34} = 4.45 \times 10^{-8} \times [OH^{-}]^{2}$

$[OH^{-}] = 6.7 \times 10^{-4}$ M

Also,

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{6.7 \times 10^{-4}}$

= $1.49 \times 10^{-11} M$

Relation between pH and concentration of hydrogen ions is as follows.

pH = $-log [H^{+}]$

= $-log (1.49 \times 10^{-11} M)$

= 10.82

Thus, we can conclude that a minimum pH of 10.82 is necessary to reduce the dissolved cadmium ion concentration to the standard.

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3 years ago

Convert 1000dm to km

polet [3.4K]

1000dm = 0.1 km

The picture below is proof that it is correct.

5 0

3 years ago

Read 2 more answers

blsea [12.9K]

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3 years ago

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago