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2 years ago

An atom with5 protons 6 neutrons and 5 electrons has an atomic mass of Amy

1 answer:

8 0

Answer: The number of neutrons for a given element is the only number that can change and still have the identity of the element stay the same, (because the atomic number is the number of protons). In this case the mass number would be 11.

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A hydrogen-filled balloon was ignited and 2.10 g of hydrogen reacted with 16.8 g of oxygen. How many grams of water vapor were f

stiks02 [169]

Hydrogen + oxygen --> water
2,1g + 16,8g = x
x = 18,9g

5 0

2 years ago

11. Fill in the blanks below for a general chemical reaction (products, reactants).

Bad White [126]

Reactants on the left and products on the right

5 0

3 years ago

An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

Anon25 [30]

Answer: The molar mass of unknown triprotic acid is 97.66 g/mol

Explanation:

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of triprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL$

Putting values in above equation, we get:

$3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

$0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol$

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0

2 years ago

Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon

balu736 [363]

Answer:

In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

In 1772, Joseph Priestley in his Recherches sur diversces especes d'air differentiated the air of animal respiration from that emitted by vegetables in the presence of light. Of the latter, which he called "dephlogistic air", he highlighted his purifying property of the environment indicating that: plants far from affecting the air in the same way as animal respiration, produce the opposite effects, and tend to preserve the sweet and healthy atmosphere , when it becomes harmful as a result of the life and breathing of the animals or their death and their rot.

In 1780, Jean Ingeshousz in his Experiences sur les vegetaux completed and reaffirmed the observations of Joseph Priestley. At the same time, he could deny Charles Bonnet's hypothesis, by demonstrating that the air expelled from the leaves comes from inside, and that the stimulating factor of the gaseous emission was not the heat produced by the sun, but the intensity of the light .

It was, finally, Jean Senebier that between 1782 and 1784, found that the "fixed air" dissolved in the water favors the vegetation. From these observations, he hypothesized that "fixed air" (carbon dioxide) is absorbed by the plants, which take it from the atmosphere with the humidity it has and in which it is mixed. Once this gas has been captured, both from the atmosphere and from the ground, it is decomposed in the presence of light by the leaves, releasing the "vital air" (oxygen) and leaving the carbon in the plant.

Thus, at the end of the century the participation of the atmosphere in plant dynamics was already seated, although the how and why of this participation were still unknown and no theory had been formulated to explain the nutritional process as a whole.

3 0

3 years ago

How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]

lisabon 2012 [21]

Balance the equation first:

2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 = 110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce = $\frac{3}{2}$ x moles of KClO3

So, O2 = (3 / 2) x 0.89

= 1.34 moles

So, Volume at STP = nRT / P

T = 273.15 K
P = 1 atm

So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L

3 0

3 years ago