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Svetradugi [14.3K]
2 years ago
7

How many particles would be found in a 12.7g sample of ammonium carbonate

Chemistry
1 answer:
Semenov [28]2 years ago
4 0

Answer: The sample of ammonium carbonate contains

0.560 mol NH

4

Explanation:The chemical formula for ammonium carbonate is

(

NH

4

)

2

CO

3

. The formula indicates that in one mole of ammonium carbonate, there are two moles of ammonium ions,

NH

4

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How much water(in grams) at its boiling point can be vaporized by adding 1.50 kJ of heat? The molar heat of vaporization for wat
LekaFEV [45]

Answer:

0.66g of water

Explanation:

Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.

If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:

1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.

As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:

0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>

6 0
2 years ago
Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod
son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0
2 years ago
The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 2.69 × 1012 at 25°c. calculate e o for a cell m
Mariulka [41]
Sr(s)+Mg²+(aq)→Sr²+(aq)+Mg(s)
Number of e-'s transfered are, n=2. Equilibrium constant,
K=2.69×10∧12
ΔG=-2.303RT logK
R=gasconstant=8.314J/mol-k
T= temperature in K= 25 oC=25+273=298K
The value we get ΔG = -70922.3J. But ΔG = -nFE
n= number of e-'s transfered in the reaction =2
F= farady = 96500C
E=potential of the cell is what?
∴E = ΔG.nF
=-(-70922.3)/2×96500)
=0.367v.
8 0
2 years ago
Which of the following statements concerning hydrocarbons is/are correct?
Alona [7]

Answer:

1.  Saturated hydrocarbons may be cyclic or acyclic molecules.

2.  An unsaturated hydrocarbon molecule contains at least one double bond.

Explanation:

Hello,

In this case, hydrocarbons are defined as the simplest organic compounds containing both carbon and hydrogen only, for that reason we can immediately discard the third statement as ethylenediamine is classified as an amine (organic chain containing NH groups).

Next, as saturated hydrocarbons only show single carbon-to-carbon bonds and carbon-to-hydrogen bonds, they may be cyclic (ring-like-shaped) or acyclic (not forming rings), so first statement is true

Finally, since we can find saturated hydrocarbons which have single carbon-to-carbon and carbon-to-hydrogen bonds only and unsaturated hydrocarbons which could have double or triple bonds between carbons and carbon-to-hydrogen bonds, the presence of at least one double bond makes the hydrocarbon unsaturated.

Therefore, first and second statements are correct.

Best regards.

6 0
2 years ago
How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m
Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})  

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

[CH_3COONa]=?mol  

[CH_3COOH]=0.200mol

pH = 5.00

Putting values in above equation, we get:

5=4.74+\log(\frac{[CH_3COONa]}{0.200})

[CH_3COONa]=0.364mol

To calculate the mass of sodium acetate for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0
2 years ago
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