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Paraphin [41]
3 years ago
14

Please help I have a test

Chemistry
1 answer:
natka813 [3]3 years ago
8 0

Answer:

Pink- Lithium Salts or Lithium Nitrate (NO3), Yellow- Sodium Salts or Sodium Nitrate (NO3)

Explanation:

These chemicals (Sr, Ba, Cu, Ca, Na, Mg) are used because they create the colors for the firework. As you can see, the chemicals Nitrogen (N) and Oxygen (O) are combined with the color chemicals to make them combust.

The chart below explains the colorants used in firework creations:

<u>Color</u>= <u>Compound</u>

Red=       strontium salts, lithium salts

lithium carbonate, Li2CO3 = red

strontium carbonate, SrCO3 = bright red

Orange= calcium salts

calcium chloride, CaCl2

calcium sulfate, CaSO4·xH2O, where x = 0,2,3,5

Gold= incandescence of iron (with carbon), charcoal, or lampblack

Yellow= sodium compounds

sodium nitrate, NaNO3

cryolite, Na3AlF6

Electric White= white-hot metal, such as magnesium or aluminum

barium oxide, BaO

Green= barium compounds + chlorine producer

barium chloride, BaCl+ = bright green

Blue= copper compounds + chlorine producer

copper acetoarsenite (Paris Green), Cu3As2O3Cu(C2H3O2)2 = blue

copper (I) chloride, CuCl = turquoise blue

Purple= mixture of strontium (red) and copper (blue) compounds

Silver= burning aluminum, titanium, or magnesium powder or flakes

I hope this answers the question you have! If not, I apologize.

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Ostrovityanka [42]

Answer:

Thats right! Gj!

Explanation:

4 0
3 years ago
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What are the representative elements usually called
antiseptic1488 [7]

Answer:

There are two sets of groups in the periodic table. The first set are Group A elements and are also known as representative elements. The second set are Group B elements and are also known as transition metals. Representative elements are the most abundant elements on earth.

4 0
3 years ago
What are the main reasons that organic chemistry is a primarily target for "green chemistry"? I. Organic solvents are often flam
Debora [2.8K]

Answer: Option 1 and 11

1. Organic solvents are often flammable.

II. Organic solvents are often toxic.

Explanation:

Organic solvents are often flammable and Organic solvents are often toxic are the reasons why organic chemistry is a primary target of green chemistry because green chemistry is the use of principles or approach to remove hazardous or toxic substances from chemical.profuctd or processes to make them fit or safe for use and the solvents of organic chemist are toxic, therefore green chemistry remove the toxic substances.

8 0
3 years ago
Need help with 14 and 16 pls asap!! this is my friends test and im taking it tomorrow!!
marin [14]

Answer:

Q14: 17,140 g = 17.14 kg.

Q16: 504 J.

Explanation:

<u><em>Q14:</em></u>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).

m is the mass of the ice (m = ??? g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).

∵ Q = m.c.ΔT

∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)

∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.

<u><em>Q16:</em></u>

  • To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat absorbed by ice (Q = ??? J).

m is the mass of the ice (m = 12.0 g).

c is the specific heat of the ice (c of ice = 2.1 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).

∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.

6 0
3 years ago
A gas contains 75.0 wt% methane, 10.0% ethane, 5.0% ethylene, and the balance water. (a) Calculate the molar composition of this
NeTakaya

Answer:

a)  molar composition of this gas on both a wet and a dry basis are

5.76 moles and 5.20 moles respectively.

Ratio of moles of water to the moles of dry gas =0.108 moles

b) Total air required = 68.51 kmoles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

Explanation:

Let assume we have 100 g of mixture of gas:

Given that :

Mass of methane =75 g

Mass of ethane = 10 g

Mass of ethylene = 5 g

∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)

Their molar composition can be calculated as follows:

Molar mass of methane CH_4}= 16 g/mol

Molar mass of ethane C_2H_6= 30 g/mol

Molar mass of ethylene C_2H_4 = 28 g/mol

Molar mass of water H_2O=18g/mol

number of moles = \frac{mass}{molar mass}

Their molar composition can be calculated as follows:

n_{CH_4}= \frac{75}{16}

n_{CH_4}= 4.69 moles

n_{C_2H_6} = \frac{10}{30}

n_{C_2H_6} = 0.33 moles

n_{C_2H_4} = \frac{5}{28}

n_{C_2H_4} = 0.18 moles

n_{H_2O}= \frac{10}{18}

n_{H_2O}= 0.56 moles

Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles

= 5.76 moles

Total moles of gas for dry basis = (5.76 - 0.56)moles

= 5.20 moles

Ratio of moles of water to the moles of dry gas = \frac{n_{H_2O}}{n_{drygas}}

= \frac{0.56}{5.2}

= 0.108 moles

b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:

    CH_4 + 2O_2_{(g)} ------> CO_2_{(g)} +2H_2O

4.69         2× 4.69

moles       moles

   C_2H_6+ \frac{7}{2}O_2_{(g)} ------> 2CO_2_{(g)} + 3H_2O

0.33      3.5 × 0.33

moles    moles

    C_2H_4+3O_2_{(g)} ----->2CO_2+2H_2O

0.18           3× 0.18

moles        moles

Mass flow rate = 100 kg/h

Their Molar Flow rate is as follows;

CH_4 = 4.69 k moles/h\\C_2H_6 = 0.33 k moles/h\\C_2H_4=0.18kmoles/h

Total moles of O_2 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles

= 11.075 k moles.

In 1 mole air = 0.21 moles O_2

Thus, moles of air required = \frac{1}{0.21}*11.075

= 52.7 k mole

30% excess air = 0.3 × 52.7 k moles

= 15.81 k moles

Total air required = (52.7 + 15.81 ) k moles/h

= 68.51 k moles/h

So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.

5 0
4 years ago
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