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3 years ago

Electron configuration of Mo

1 answer:

3 0

Answer:The computer writes electron configurations in the form, for Co for example, [Ar]3d(7)4s(2), which we would write [Ar]3d74s2

NOTE: Write electron configurations as they appear in the periodic table in the front of you text put 4f before 5d

Explanation:

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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

natima [27]

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given : 1) Molarity of Na⁺ ions = 1.00 M or 1.00 $\frac{mol}{L}$

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL

$Volume of solution = 725 mL * \frac{1L }{1000 mL}$

Volume of solution = 0.725 L

Following steps can be done to find mass of :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

$Na_3PO_4 3 Na^+ + PO_4^3^-$

Step 2: Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula which is : $Molarity (\frac{mol}{L} ) = \frac{mole of Na^+ }{volume of solution }$

Plugging value of Molarity and volume

$1.00 \frac{mol}{L} = \frac{Mole of Na^+ ions}{0.725 L}$

Multiplying both side by 0.725 L

$1.00 \frac{mol}{L}* 0.725 L = \frac{mole of Na^+ ions}{0.725 L} * 0.725 L$

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion = 3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

$Mole of Na_3PO_4= Mole of Na^+ * Mole ratio$

$Mole of Na_3PO_4 = 0.725 mol * \frac{1 mole of Na_3PO_4}{3 mole of Na^+ }$

Mole of Na₃PO₄ = 0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using molar mass of Na₃PO₄ as :

$Mass (g) = mole (mol) * molar mass \frac{g}{mol}$

$Mass of Na_3PO_4 = 0.242 mol * 163.94 \frac{g}{mol}$

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8 0

3 years ago

What are properties or characteristics shared by Metalloids and Nonmetals?

Mekhanik [1.2K]

Nonmetals have properties opposite those of the metals. The nonmetals are brittle, not malleable or ductile, poor conductors of both heat and electricity, and tend to gain electrons in chemical reactions. Some nonmetals are liquids. These elements are shown in the following figure.

3 0

3 years ago

Draw the addition product formed when one equivalent of hcl reacts with the following diene.

pashok25 [27]

Answer:

Major Product = 4-chloro-4-methylcyclohex-1-ene

Explanation:

Alkene are the class of organic compounds which contain one or more double bonds between two carbon atoms. Alkenes are considered most reactive among the unsaturated hydrocarbons and they undergo addition reactions due to high electron density around the double bonds.

In given question it is written that we are provided with one equivalent of HCl while, our compound contains two double bonds (diene) so in selected starting material the HCl will be added across (hydrohalogenation reaction) the substituted double bond because it will give a more stable carbocation (tertiary carbocation) during the reaction course. Hence, as shown in reaction scheme 4-chloro-4-methylcyclohex-1-ene will be the major product.

5 0

3 years ago

Please help im stuck

baherus [9]

Answer:

Spring

Explanation:

4 0

3 years ago

How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?

allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

volume Na2S = 0.0212 L = 21.2 mL

6 0

3 years ago