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Katena32 [7]
3 years ago
12

Is Li has 7 valence electrons

Chemistry
2 answers:
Bumek [7]3 years ago
7 0
Lithium's valence electron configuration shell is actually 2s1. Lithium has 1 valence electron (outer most shell).
jeka943 years ago
3 0
This is false. The amount of electrons in lithium (Li) is not 7, it is 3. There are 3 electrons in lithium.
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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
san4es73 [151]

Answer:

83.20 g of Na3PO4

Explanation:

1 mole of Na3PO4 contains 3 moles of Na+.

Mole of Na ion to be prepared = Molarity x volume

                 = 0.700 x 725/1000

                     = 0.5075 mole

If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:

           0.5075/3 x 1 = 0.1692 mole of Na3PO4

mole of Na3PO4 = mass/molar mass = 0.1692

Hence, mass of Na3PO4 = 0.1692 x molar mass

                                     = 0.1692 x 163.94

                                        = 83.20 g.

83.20 g of Na3PO4 will be needed.

3 0
3 years ago
Read 2 more answers
Part A
KonstantinChe [14]

Answer:

The answer to your question is below

Explanation:

A.

[H₃O⁺] = 2 x 10⁻¹⁴ M

pH = ?

Formula

                                    pH = - log [H₃O⁺]

Substitution

                                    pH = - log [2 x 10⁻¹⁴]

Result

                                    pH = 13.7          

B.

[H₃O⁺] = ?

pH = 3.12

Formula

                                   pH = - log [H₃O⁺]

Substitution

                                   3.12 = - log [H₃O⁺]

                                   10^{-3.12} = [H_{3} O^{+}]

Result

                                  [H₃O⁺] = 7.59 M

   

7 0
3 years ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
mamaluj [8]

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

2SO_2 + O_2 \to 2SO_3

The I.C.E table can be represented as:

                     2SO₂              O₂                   2SO₃

Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,  

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction 2SO_2 + O_2 \to 2SO_3 = 0.8325;

If we want to find:

SO_2 + \dfrac{1}{2}O_2 \to SO_3

Then:

K_c = (0.8325)^{1/2}

\mathbf{K_c = 0.912}

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

7 0
3 years ago
Desiré was investigating a chemical reaction.When she heated it up, she found that sulfuric acid changed into water. She made th
pochemuha

Answer:

no it is not a complete model

3 0
3 years ago
Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT
Andru [333]

Answer:

6.14\cdot 10^{-6}

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

K_{eq} = \frac{[ADP][Pi]}{ATP}

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

\Delta G^o = -RT ln K_{eq}

From here, rearrange the equation to solve for K:

K_{eq} = e^{-\frac{\Delta G^o}{RT}}

Now we know from the initial equation that:

K_{eq} = \frac{[ADP][Pi]}{ATP}

Let's express the ratio of ADP to ATP:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}

Substitute the expression for K:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}

Now we may use the values given to solve:

\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}

7 0
3 years ago
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