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3 years ago

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When comparing land and sea temperatures, the temperatures of the sea are more constant than the temperatures of the land throug

hout the day
Which of the following explains why?

A The water loses more heat than it gains

B The water must gain or lose more heat than the land to have the same change in temperatures

C The water does not absorb or lose heat quickly

D The water loses heat faster than the land does.

2 answers:

EleoNora [17]3 years ago

6 0

Answer: it's b

Explanation:Land gains heat faster than water therefore water has to gain more heat to be constant

eimsori [14]3 years ago

4 0

The answer to it is (b)

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A solution of Na2CO3 is added dropwise to a solution that contains 1.15×10−2 M Fe2+ and 0.58×10−2 M Cd2+. What concentration of

castortr0y [4]

The question is incomplete, complete question is;

A solution of $Na_2CO_3$ is added dropwise to a solution that contains $1.15\times 10^{-2} M$ of $Fe^{2+}$ and $0.58\times 10^{-2} M$ and $Cd^{2+}$ .

What concentration of $CO_3^{2-}$ is need to initiate precipitation? Neglect any volume changes during the addition.

$K_{sp}$ value $FeCO_3: 2.10\times 10^{-11}$

$K_{sp}$ value $CdCO_3: 1.80\times 10^{-14}$

What concentration of $CO_3^{2-}$ is need to initiate precipitation of the first ion.

Answer:

Cadmium carbonate will precipitate out first.

Concentration of $CO_3^{2-}$ is need to initiate precipitation of the cadmium (II) ion is $3.103\times 10^{-12} M$ .

Explanation:

1) $FeCO_3\rightleftharpoons Fe^{2+}+CO_3^{2-}$

The expression of an solubility product of iron(II) carbonate :

$K_{sp}=[Fe^{2+}][CO_3^{2-}]$

$2.10\times 10^{-11}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{2.10\times 10^{-11}}{1.15\times 10^{-2} M}$

$[CO_3^{2-}]=1.826\times 10^{-9}M$

2) $CdCO_3\rightleftharpoons Cd^{2+}+CO_3^{2-}$

The expression of an solubility product of cadmium(II) carbonate :

$K_{sp}=[Cd^{2+}][CO_3^{2-}]$

$1.80\times 10^{-14}=0.58\times 10^{-2} M\times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{1.80\times 10^{-14}}{0.58\times 10^{-2} M}$

$[CO_3^{2-}]=3.103\times 10^{-12} M$

On comparing the concentrations of carbonate ions for both metallic ions, we can see that concentration to precipitate out the cadmium (II) carbonate from the solution is less than concentration to precipitate out the iron (II) carbonate from the solution.

So, cadmium carbonate will precipitate out first.

And the concentration of carbonate ions to start the precipitation of cadmium carbonate we will need concentration of carbonate ions greater than the $3.103\times 10^{-12} M$ concentration.

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How are strong acids and strong bases similar?

Lady_Fox [76]

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An acid or base which strongly conducts electricity contains a large number of ions and is called a strong acid or base and an acid or base which conducts electricity only weakly contains only a few ions and is called a weak acid or base.

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Which of the solids in the following set should you test to investigate for the effects of ANIONS on pH? Will give brainliest

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Answer:

KCIO

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2 years ago

The notation for a particular nucleus is r3785b. in an electrically neutral atom. How many electrons are in orbit about this nuc

Serhud [2]

$^{85}_{37}Rb$ has 37 electrons in the orbit about its nucleus.

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Atomic mass is the total number of neutrons and protons present in the nucleus of an element.

The atomic number is the total number of protons present in the nucleus of an element.

For writing an atomic symbol, atomic mass is written on the top left of the atomic symbol, and the atomic number is written on the downside of the left of the symbol.

$^{Atomic mass}_{Atomic number}X$

An electrically neutral atom is an atom that has an equal number of electrons as well as protons.

Thus from the above conclusion we can say that $^{85}_{37}Rb$ has 37 electrons.

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1 year ago