1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mezya [45]
4 years ago
8

Here are some data from a similar experiment, to determine the empirical formula of an oxide of tin. Calculate the empirical for

mula according to these data. mass of crucible, cover, and tin sample 21.76 g mass of empty crucible with cover 19.66 g mass of crucible and cover and sample, after prolonged heating gives constant weight 22.29 g
Chemistry
1 answer:
eduard4 years ago
7 0

Answer:

Empirical formula of the Tin oxide sample is SnO₂

Explanation:

Tin reacts with combines with oxygen to form an oxide of tin.

Mass of crucible with cover = 19.66 g

Mass of crucible, cover, and tin sample = 22.29 g

Mass of crucible and cover and sample, after prolonged heating gives constant weight = 21.76 g

Mass of Tin oxide sample = 22.29 - 19.66 = 2.63 g

Mass of ordinary tin, after heating to breakdown the tin and oxygen = 21.76 - 19.66 = 2.1 g

Meaning that, mass of oxygen in the tin oxide sample = 2.63 - 2.1 = 0.53 g

Mass of Tin in the Tin Oxide sample = 2.1 g

Mass of Oxygen in the Tin oxide sample = 0.53 g

Convert these to number of moles

Number of moles of Tin on the Tin oxide sample = 2.1/118.71 = 0.0177

Number of moles of Oxygen in the Tin oxide sample = 0.53/16 = 0.0335

divide the number of moles by the lowest number

0.0177:0.0335

It becomes,

1:2

SnO₂

Hence, the empirical formula for the Tin oxide sample = SnO₂

You might be interested in
A neutralization reaction between an acid an sodium hydroxide formed water and the salt named sodium sulfate. What was the formu
PSYCHO15rus [73]
Let's see how a neutralization reaction produces both water and a salt, using as an example the reaction between solutions of hydrochloric acid and sodium hydroxide. The overall equation for this reaction is: NaOH + HCl → H2O and NaCl. Now let's break this reaction down into two parts to see how each product forms.
4 0
3 years ago
A 1.00-L gas sample at 100.°C and 500. torr contains 52.0% helium and 48.0% xenon by mass. What are the partial pressures of the
Nataly_w [17]

Answer:

Partial pressure of He=486 torr, partial pressure of Xe= 14 torr

Explanation:

Using the equation, PV=nRT----------------------------------------(1)

Making n the subject of the formula;

n= PV/RT---------------------(2)

Where n= number of moles, v= volume, T= temperature, P= volume.

n= (500 torr/760 torr × 1 atm)× 1L ÷ 373K ×0.082 L atmK^-1. Mol^-1

n= 0.6579 atm.L/ 30.6233

n= 0.0215 mol.

Let the total mass of the gas= 2b(in g).

Mass of helium gas = b (in g) = mass of Xenon gas

Mole of helium gas= b(in g) / 4 gmol^-1

=b/4 mol

Mole of Xenon= b g/131.3 gmol^-1

= b/131.3 mol.

Solving for b, we have;

b/4+b/131.3 = 0.0215 mole

(131.3+4)b/525.2= 0.0215

Multiply both sides by 1/135.3.

b= 11.2918/135.3

b= 0.0835 g

Mole of He gas= 0.0835/4= 0.0209

Mole of Xe gas= 0.0215- 0.0209

= 0.0006 mol

Mole fraction of He = 0.0209/0.0215

= 0.972

Mole fraction of Xe= 0.0006/0.0215

= 0.028

Partial pressure of He gas= 0.972× 500 torr= 486 torr

Partial pressure of Xe gas= 0.028 ×500 torr = 14 torr

4 0
4 years ago
2Na2O2 + 2CO2 → 2Na2CO3 + O2
ahrayia [7]

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield  x 100%

81.0  % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

7 0
3 years ago
Read 2 more answers
An isotope with a half life of 10 min and a starting sample size of 100 g would have
IgorC [24]

Answer:

12.5 gm left

Explanation:

30 minutes is THREE half lives

   1/2^3 = 1/8th   would be left

100g * 1/8 = 12.5 gm

7 0
2 years ago
What do covalent bonds usually form at room temperature?.
hichkok12 [17]

Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation.

7 0
3 years ago
Other questions:
  • How many compounds can i make with Sodium Calcium Hydrogen oxygen Carbon and chlorine
    13·1 answer
  • On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles conta
    10·1 answer
  • Using the enthalpy of reaction for two reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine wit
    15·1 answer
  • The size of the gray circle represents atomic radius, or
    14·1 answer
  • When standardizing an iodine solution, 30.37 mL of the solution was required to
    10·1 answer
  • 2 upper N upper o upper c l double-headed arrow 2 upper N upper O (g) plus upper C l subscript 2 (g).
    5·1 answer
  • How many moles are there in 36.0 grams of water?
    11·1 answer
  • Which of the following statements is related to a postulate (assumption) of the kinetic molecular theory of gases?
    12·1 answer
  • Advantages and disadvantages of Law of Triad :
    13·1 answer
  • Katie wants to do an experiment to test what factors affect the growth rate of sunflowers. She plants two groups of sunflower se
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!