Scientists often look at how much isotopes of _____

Which of the following is true about covalent bonding?

zhenek [66]

Answer:

Electrons are shared in pairs!

Explanation:

I'm doing learning about Covalent Bonding too:)

7 0

3 years ago

Calculate the work (kJ) done during a reaction in which the internal volume expands from 18 L to 49 L against a vacuum (an outsi

IgorLugansk [536]

Answer:

b

Explanation:

b

4 0

3 years ago

How many moless is 4.91x10^22 molecules of H3PO4?

Lynna [10]

1 mol = 6.023x10^23 number of molecules (Avogadro's number)

1 : 6.023x10^23
X : 4.91x10^22

(6.023x10^23)X = 4.91x10^22

X = 4.91x10^22/6.023x10^23

X = 0.082 Moles

6 0

4 years ago

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

5 0

3 years ago

In an extraction experiment similar to your expt. 4, you attempt to separate 2-naphthol from cyclohexane, both of which are diss

Viktor [21]

Answer:

[a]. 108 mL.

[b].

1. use HCL in washing for the neutralization of NaOH.

2. Wash with water.

3. make use of MgSO4 to remove water traces.

4. Evaporate to get the pure cyclohexane

Explanation:

The chemical equation for the reaction between 2-naphthol and Sodium hydroxide is given below as:

C₁₀H₈O + NaOH --------------------------------------------------> C10H7O⁻ Na⁺ + H₂O.

The C10H7O⁻ Na⁺ is the aqueous phase.

The molarity of Sodium Hydroxide is = [ (mass of sodium hydroxide ) ÷ molar mass sodium hydroxide × volume] × 1000.

The molarity of sodium hydroxide = [ (5/40) × 100] × 1000 = 1.25 M.

The number of moles of 2-naphthol = mass / molar mass = 20 / 144.17 = 0.138 moles.

Recall that, the formula for the number of moles = concentration × volume. Therefore, the volume of sodium hydroxide is given below as:

Volume of sodium hydroxide = the number of moles of sodium hydroxide ÷ concentration of sodium hydroxide = 0.138 ÷ 1.25 = 0.108L = 108 mL.

In order to efficiently isolate relatively pure cyclohexane from the ether layer, the following process must be followed:

1. use HCL in washing for the neutralization of NaOH.

2. Wash with water.

3. make use of MgSO4 to remove water traces.

4. Evaporate to get the pure cyclohexane

7 0

3 years ago

Scientists often look at how much isotopes of ______