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2 years ago

What are the coefficients that will balance this skeleton equation? AlCl3 + NaOH → Al(OH)3 + NaCl

Chemistry

1 answer:

almond37 [142]2 years ago

3 0

Answer:

The coefficients are 1, 3, 1, 3

Explanation:

The equation;

AlCl3 + NaOH → Al(OH)3 + NaCl

can be balanced by using the coefficients 1, 3, 1, 3, such that the balanced equation will be;

AlCl3 + 3NaOH → Al(OH)3 + 3NaCl

Chemical equations needs to be balanced so as to follow the law of conservation of mass. This occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side.

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When elemental sodium is added to water, the sodium atoms ionize spontaneously. uncharged na becomes na+. this means that the na

Temka [501]

Ionized, become charged, become a cation

hopefully that helps

5 0

2 years ago

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomas

jolli1 [7]

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

Where:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

7 0

2 years ago

Which statement best describes how an ionic bond forms?

maxonik [38]

The question says,'which statement best describes how an ionic bond forms. The correct option is A. Ionic bonds are formed as a result of complete transfer of electrovalence electrons from one atom to another. The atom that donate the electron become a positively charged ion while the atom that received the atom become a negatively charged ion.

4 0

2 years ago

Read 2 more answers

The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =

Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) (1)

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] (2)

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) = 9.744g of monosodium succinate.

Mass of disodium succinate:

0.0304moles * (162g / 1mol) = 4.925g of disodium succinate.

5 0

2 years ago

A student assistant is cleaning up after a chemistry laboratory exercise and finds three one-liter bottles containing alcohol so

Svetllana [295]

The approximate alcohol content is 210 ml.

Explanation:

It can be deduced from the question that each bottle is of 1000ml or 1 litre.

The first bottle is one half full means it has 500 ml of solution and it has 20% alcohol in it. So volume of alcohol in the solution is

20/100*500

=100 ml

The first bottle is one fifth full, so the volume of mixture is 1/5th of 1000ml

so it is 200ml having 30% alcohol

30/100*200

= 60 ml

The third bottle is one tenth full so its volume is 1/10*1000

100 ml. having 50% of alcohol

50/100*100

50 ml.

The alcohol content obtained from all these 3 litres is:

100+60+50

= 210 ml of alchohol is obtained from 800 ml of mixture.

3 0

2 years ago