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Lubov Fominskaja [6]

2 years ago

5

Example of gases collected through upward delivery

2 answers:

taurus [48]2 years ago

8 0

Answer:

Gases which are collected by this process are Hydrogen, ammonia etc. In this process The more dense gases are collected in a gas jar. Carbon dioxide, Nitrogen dioxide are collected by this way. Air is displaced upward thus, this process is known as upward delivery...!

fgiga [73]2 years ago

4 0

Answer:

In this process The more dense gases are collected in a gas jar. Carbon dioxide, Nitrogen dioxide are collected by this way. Air is displaced upward thus, this process is known as upward delivery

Hope it helped :)

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A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

2 years ago

If Marie and Calvin dissolve 50 grams of KBr in 100 grams of water at 90oC, the solution is

hram777 [196]

Answer:

The solution is 50 %wt

Explanation:

50% wt is a sort of concentration and means, that 50 g of solute (in this case, the potassium bromide) dissolved in 100 g of water.

It is the same to say, that there are 50g of KBr for every 100g of H₂O

8 0

2 years ago

Diamond is an allotrope of :

Marat540 [252]

Answer:

carbon

Explanation:

because it is an allotrope of carbon

5 0

2 years ago

Read 2 more answers

An instrument that is used to measure air pressure is a ?

Zolol [24]

A barometer measures atmospheric pressure.

8 0

2 years ago

In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i

GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: $V_{1}$ = 21.35 mL, $M_{1}$ = 0.150 M

$V_{2}$ = 25.0 mL, $M_{2}$ = ?

Formula used to calculate molarity of $H_{2}SO_{4}$ is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M$

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = $V_{1} + V_{2}$

= 21.35 mL + 25.0 mL

= 46.36 mL (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol$

Thus, we can conclude that there are 0.006 moles of acid in the flask.

3 0

2 years ago