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3 years ago

Match the following scientist with his experiment or contribution

Chemistry

1 answer:

Verdich [7]3 years ago

3 0

J. J. Thomson - oil drop

Albert Einstein - atomic theory

Niels Bohr - model of the atom

Ernest Rutherford - gold foil experiment

John Dalton - cathode ray tube

Robert Millikan - photoelectric effect

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3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O

AysviL [449]

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cu: 3 moles
HNO₃: 8 moles
Cu(NO₃)₂: 3 mole
NO: 2 moles
H₂O: 4 moles

The molar mass of the compounds is:

Cu: 63.55 g/mole
HNO₃: 63 g/mole
Cu(NO₃)₂: 187.55 g/mole
NO: 30 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cu: 3 moles ×63.55 g/mole= 190.65 grams
HNO₃: 8 moles ×63 g/mole= 504 grams
Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
NO: 2 moles ×30 g/mole= 60 grams
H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

$mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3} x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }$

<u><em>mass of Cu(NO₃)₂= 5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

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7 0

2 years ago

Which symbol represents a salt?<br><br> C2H2<br><br><br> NaCl<br><br><br> O2<br><br><br> C6H12O6

True [87]

Answer:

NaCl

Explanation:

represents table salt

3 0

3 years ago

Read 2 more answers

Bromine pentafluoride BrF5 is a good example of inter halogen molecule

Snezhnost [94]

Answer:

a) The lewis dot structure is shown in the image attached to this answer

b) The formal charge on each of the atoms is zero

c) bromine has an oxidation state of +5 while fluorine has an oxidation state of -1

d) 90 degrees

e) Square Pyramidal

f) polar bonds

g) polar molecule

Explanation:

The molecule BrF5 has a formal charge of zero. It exhibits an sp3d2 hybridization state with a square pyramidal geometry. The bond angle in the molecule is 90 degrees. It is a molecule of the type AX5E. The oxidation state of bromine is +5 while that of fluorine is -1.

The Br-F bonds are polar. The overall molecule is polar due to asymmetric charge distribution concentrating on the central atom since the molecule is square pyramidal.

3 0

3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

8 0

3 years ago

Assume after recording the mass of KHP on the weigh paper it was not completely transferred into the Flask. Your following titra

maxonik [38]

Answer:

unchanged

Explanation:

KHP is potassium hydrogen phthalate. It is a weak acid. It forms a colourless ionic weakly acidic solution which can be titrated against NaOH to determine the concentration of a sodium hydroxide solution. KHP is used as a primary standard. Its concentration depends on the mass of solid added and the volume of solution.

If all the KHP was not transferred into the flask, the experimental value of the concentration of NaOH remains unchanged as long as the titration was carried out correctly and following normal procedure. Only the concentration of the KHP will be affected by this error during calculation.

5 0

3 years ago