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2 years ago

Entir What is the striking back of the bunsen- burner?

Chemistry

1 answer:

svp [43]2 years ago

6 0

Answer:

With too little air, the gas mixture will not burn completely and will form tiny carbon particles that are heated to glowing, making the flame luminous. With too much air, the flame may burn inside the burner tube; that is, it may strike back.

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This process is named taxonomy.

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3 years ago

Hi, can someone help me with this.

AleksAgata [21]

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3 years ago

How does an unconformity affect our geological knowledge of an area?

andriy [413]

Answer:

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2 years ago

Read 2 more answers

A 20.7167 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation mgco3(s) → mgo(s

marishachu [46]

The mass of impure $MgCO_{3}$ is 20.7167 g. The decomposition reaction is as follows:

$MgCO_{3}(s)\rightarrow MgO(s)+CO_{2}(g)$

Here, 1 mole of $MgCO_{3}$ gives 1 mole of MgO.

Molar mass of $MgCO_{3}$ and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

$n=\frac{m}{M}$

Here, M is molar mass.

Since, $n_{MgCO_{3}}=n_{MgO}$

Thus, $\frac{m_{MgCO_{3}}}{M_{MgCO_{3}}}=\frac{m_{MgO}}{M_{MgO}}$

On putting the values,

$\frac{m_{MgCO_{3}}}{84.31}=\frac{m_{MgO}}{40.3044}$

Rearranging,

$m_{MgO}=\frac{m_{MgCO_{3}}}{84.31}\times 40.3044$

Or,

$m_{MgCO_{3}}=\frac{m_{MgO}}{0.4780}$ ...... (1)

Let the mass of impurity be $M_{I}$ and mass of impure $MgCO_{3}$ is 20.7167 g thus,

$M_{MgCO_{3}}=(20.7167-M_{I})g$ ...... (2)

Also, mass of impure MgO is 16.8817 g thus,

$M_{MgO}=(16.8817-M_{I})g$ ...... (3)

On comparing equations (2) and (3),

$M_{MgO}=M_{MgCO_{3}}-3.835$

Putting the value of $M_{MgO}$ in equation (1),

$m_{MgCO_{3}}=\frac{M_{MgCO_{3}}-3.835}{0.4780}$

Or,

$0.4780 M_{MgCO_{3}}=M_{MgCO_{3}}-3.835$

Or,

$M_{MgCO_{3}}-0.4780M_{MgCO_{3}}=3.835$

Or,

$0.522 M_{MgCO_{3}}=3.835$

Or,

$M_{MgCO_{3}}=\frac{3.835}{0.522}=7.35 g$

Thus, magnesium carbonate present in the original sample is 7.35 g.

5 0

3 years ago

A sample of gas has a mass of 14.2g and occupies a volume of 678ml at 25.0 c and a pressure of 0.80atm. What will the volume of

11Alexandr11 [23.1K]

Answer:

497 ml

Explanation:

(678 ml) x (273 / (25 + 273)) x (0.8 atm / 1 atm) = 497 ml

3 0

3 years ago