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3 years ago

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Cabbage juice turns pink in solutions with pH below 7 and green in solutions with pH above 7. You drop some cabbage juice in a s

ink full of soapy dishes and in a glass of soda. The cabbage juice is green in the soapy water and pink in the soda. Which substance is basic?

1 answer:

Ann [662]3 years ago

8 0

<span>Basic substances, when dissolved in water, give pH above 7</span>

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Can someone please quickly answer this question?

dimulka [17.4K]

Answer:

The difference between the two is:

$[Kr]4d^1^05s^25p^2$ is the Orbital Occupancy

$[Kr]5s^24d^1^05p^2$ is the Orbital Filling Order

Both are correct, I don't think your teacher will be so nit-picky to care.

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3 years ago

The effects of acid rain include:

posledela

<span>The effects of acid rain include:

d. All the above
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3 years ago

Solid NaI is slowly added to a solution that is 0.0079 M Cu+ and 0.0087 M Ag+.Which compound will begin to precipitate first?NaI

NeTakaya

Answer :

AgI should precipitate first.

The concentration of $Ag^+$ when CuI just begins to precipitate is, $6.64\times 10^{-7}M$

Percent of $Ag^+$ remains is, 0.0076 %

Explanation :

$K_{sp}$ for CuI is $1\times 10^{-12}$

$K_{sp}$ for AgI is $8.3\times 10^{-17}$

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.

Now we have to calculate the concentration of iodide ion.

The solubility equilibrium reaction will be:

$CuI\rightleftharpoons Cu^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Cu^+][I^-]$

$1\times 10^{-12}=0.0079\times [I^-]$

$[I^-]=1.25\times 10^{-10}M$

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

$AgI\rightleftharpoons Ag^++I^-$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^+][I^-]$

$8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M$

$[Ag^+]=6.64\times 10^{-7}M$

Now we have to calculate the percent of $Ag^+$ remains in solution at this point.

Percent of $Ag^+$ remains = $\frac{6.64\times 10^{-7}}{0.0087}\times 100$

Percent of $Ag^+$ remains = 0.0076 %

8 0

3 years ago

You have 500,000 atoms of a radioactive substance. After 2 half-lives have past, how many atoms remain?

Aleonysh [2.5K]

Answer:

125000

Explanation:

Because it is halved and halved again.

3 0

2 years ago

Read 2 more answers

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

<u>Convert Atoms to Moles</u>

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

<u>Convert Moles to Grams</u>

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

<u>Round</u>

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.<u>4</u>184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately <u>147 grams.</u>

3 0

3 years ago