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ololo11 [35]
3 years ago
5

1. The following are statements about flames some of them are true while others are

Chemistry
1 answer:
hoa [83]3 years ago
4 0

Answer:

i)The carbon particles are formed when the gas is not completely burnt(True)

ii)A solid or liquid must be change to gaseous state in order to make flame(TRUE)

iii)A non- luminous Bunsen flame produces has three visible regions(FALSE)

iv)The luminous Bunsen flame produces less soot than the non- luminous flame(FALSE)

v)A yellow Bunsen flame is obtained when the air holes are opened(FALSE)

vi)The luminous Bunsen flame is hotter than a non- luminous Bunsen flame(FALSE)

vii)Bunsen burner uses kerosene as a source of heat(FALSE)

Explanation:

i)When the incomplete combustion ,then CO(carbon monoxide) and C(carbon) are produced. <u>The unburnt carbon get deposited as</u><u> Soot</u>.

In complete combustion CO2(carbon dioxide) is produced.

ii)Fire, is the mixture of hot gases.Flame are result of chemical reaction between oxygen and fuel(e.g butane ,wood).Flame contain,carbon dioxide , heat ,light,steam.

iii)There are total three regions of flame in which the outer most region is the non-luminous .The non-luminous part is <u>blue(one visible region</u>) and does not emit-light.

iv) Luminous Bunsen flame produces more soot than the non- luminous flame because luminous flame contain more unburnt particles as compare to non-luminous.

luminous part has less oxygen available for complete combustion of fuel

v)When the air hole is open it means excess oxygen is supplied to the fuel.

In this case clean and luminous flame is produced.When air hole is close yellow flame is produced.

vi)Non-luminous part of flame is more hot because of complete combustion due to proper availability  of oxygen. Sometime the particle of non-luminous flame get ionize also.

vii).The bunsen burner uses Natural gas(methane) or LPG(propane and butane)as sources of heat because these are easily burnt not kerosene.

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4 0
3 years ago
Magma that is rich in feldspar and silica is called?​
weeeeeb [17]

Answer:

<h3>magma that is rich in feldspar and silica is called STRATOVOLCANO.</h3>
3 0
3 years ago
Please help me out i will give you brainlist. 0.500 is wrong
Alik [6]
<h3>Answer:  b) 0.250 mol</h3>

============================================

Work Shown:

Using the periodic table, we see that

  • 1 mole of carbon = 12 grams
  • 1 mole of oxygen = 16 grams

These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.

So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.

-------------------

Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.

11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2

In short,

11.0 grams of CO2 = 0.250 mol of CO2

This is approximate.

We don't need to use any of the information in the table.

3 0
2 years ago
Read 2 more answers
How many grams of KCl is needed to make .75 L of a 1 M solution of KCl?
Zielflug [23.3K]

Answer:

The answer to your question is 25.9 g of KCl

Explanation:

Data

Grams of KCl = ?

Volume = 0.75 l

Molarity = 1 M

Formula

Molarity = \frac{number of moles}{volume}

Solve for number of moles

Number of moles = Molarity x volume

Substitution

Number of moles = 1 x 0.75

Simplification

Number of moles = 0.75 moles

Molecular mass KCl = 39 + 35.5 = 34.5

Use proportions to find the grams of KCl

                          34.5 g of KCl ----------------  1 mol

                             x                  ----------------  0.75 moles

                            x = (0.75 x 34.5) / 1

                            x = 25.9 g of KCl

3 0
3 years ago
An experiment shows that a 236 mL gas sample has a mass of 0.443 g at a pressure of 740 mmHg and a temperature of 22 ∘C. What is
aleksley [76]

Answer:

49.2 g/mol

Explanation:

Let's first take account of what we have and convert them into the correct units.

Volume= 236 mL x (\frac{1 L}{1000 mL}) = .236 L

Pressure= 740 mm Hg x (\frac{1 atm}{760 mm Hg})= 0.97 atm

Temperature= 22C + 273= 295 K

mass= 0.443 g

Molar mass is in grams per mole, or MM= \frac{mass}{moles} or MM= \frac{m}{n}. They're all the same.

We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be

n=\frac{PV}{RT}, where R (constant)= 0.082 L atm mol-1 K-1

Let's plug in what we know.

n=\frac{(0.97 atm)(0.236 L)}{(0.082)(295K)}

n= 0.009 mol

Let's look back at MM= \frac{m}{n} and plug in what we know.

MM= \frac{0.443 g}{0.009 mol}

MM= 49.2 g/mol

3 0
2 years ago
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