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4 years ago

When a scientist shares her findings with other scientists she is

Chemistry

2 answers:

finlep [7]4 years ago

5 0

Answer:

communicating results

Explanation:

Alecsey [184]4 years ago

5 0

Answer:

Communicating results

Explanation:

You might be interested in

The individual dipole moments in ammonia (NH3) do not cancel each other

kozerog [31]

Answer:

The strongest force that exists between molecules of Ammonia is <em>Hydrogen Bonding</em>.

Explanation:

Hydrogen Bond Interactions are those interactions which are formed between a partial positive hydrogen atom bonded directly to most electronegative atoms (i.e. F, O and N) of one molecule interacts with the partial negative most electronegative atom of another molecule.

Hence, in ammonia the nitrogen atom being more electronegative element than Hydrogen will be having partial negative charge and making the hydrogen atom partial positive. Therefore, the attraction between these partials charges will be the main force of interaction between ammonia molecules.

Other than Hydrogen bonding interactions ammonia will also experience dipole-dipole attraction and London dispersion forces.

4 0

3 years ago

how could you check that when the calcium carbonate is reacting with hydrochloric acid the gas given off is carbon dioxide

pav-90 [236]

Answer:

CaCO3 + 2 HCl => CaCl2 + CO2 + H2O

Explanation:

CO2 will burst out when the reaction occurs in water, and the ions will be dissociated: Ca(2+) + 2 Cl (-)

7 0

3 years ago

You may have to recrystallize any or all of the components of your extraction mixture, benzoic acid, ethyl-4-aminobenzoate and f

hjlf

Answer:

the Recrystallization solvents are;

-Sodium Hydroxide Base (NaOH)

- Hydrochloric Acid (HCl)

- Sand bath when acidic and basic components are removed.

Explanation:

Recrystallization is simply a technique used to purify an impure compound in a solvent.

Now, we want to purify the mixture of benzoic acid, Ethyl-4-aminobenzoate and fluorenone.

For the benzoic acid, it can be separated out of the mixture by addition of sodium hydroxide base (NaOH).

The Ethyl - 4 - aminobenzoate will be separated from the mixture by the addition of Hydrochloric Acid (HCl).

The fluorenone would be separated out by heating the mixture in a sand bath after the basic and acidic components have already been extracted out.

Thus, the Recrystallization solvents are;

-Sodium Hydroxide Base (NaOH)

- Hydrochloric Acid (HCl)

- Sand bath when acidic and basic components are removed.

6 0

3 years ago

What is the pOH of a 3.0 x10-3 M NaOH solution?<br> pOH =

kodGreya [7K]

Answer:

2.5

Explanation:

From the question given above, the following data were obtained:

Molarity of NaOH = 3.0x10¯³ M

pOH =?

Next, we shall determine the concentration of the hydroxide ion in the solution. This can be obtained as follow:

NaOH (aq) —> Na⁺ (aq) + OH¯ (aq)

From the balanced equation above,

1 mole of NaOH produced 1 mole of OH¯.

Therefore, 3.0x10¯³ M NaOH will also produce 3.0x10¯³ M OH¯.

Finally, we shall determine the pOH of the solution. This can be obtained as illustrated below:

Concentration of hydroxide ion [OH¯] = 3.0x10¯³ M

pOH =?

pOH = – Log [OH¯]

pOH = – Log 3.0x10¯³

pOH = 2.5

Thus, the pOH of the solution is 2.5

4 0

3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0

3 years ago