HELP! HELP! an element with 5 protons, and 8 electrons has an atomic number of?

In the periodic table , the elements are organized in order of ________atomic number .

jok3333 [9.3K]

The correct answer is increasing

4 0

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

An excess of Ba(No3)2 reacts with 250ml of H2SO4 solution to give 0.55g of BaSo4.determine The concentration in moles per litre

Wewaii [24]

Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.

7 0

3 years ago

Please help is for today

love history [14]

Answer:

Our bodies are designed to regulate our temperature. When it's cold outside, your body makes sure to keep the blood flowing to your core and vital organs to keep them warm. This can change the amount of blood flow to your hands and feet, making them feel cold

5 0

3 years ago

68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi

Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we convert the given masses of reactants to moles, using their respective molar masses:

68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH

78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and Mg(NO₃)₂ is the limiting reagent.

Now we calculate how many Mg(OH)₂ are produced, using the moles of the limiting reagent:

0.528 mol Mg(NO₃)₂ * $\frac{1molMg(OH)_2}{1molMg(NO_3)_2}$ = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g

7 0

2 years ago