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3 years ago

What elements are in fire works?

Chemistry

1 answer:

anzhelika [568]3 years ago

4 0

Metal salts commonly used in firework displays

Red Fireworks: Strontium Ccarbonate

Orange Fireworks: Calcium Chloride

Yellow Fireworks: Sodium Nitrate

Green Fireworks: Barium Chloride

Blue Fireworks: Copper Chloride

Beside each fireworks name is what is in those fireworks.

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Mendeleev arranged the known chemical elements in a table according to increasing

vesna_86 [32]

Atomic mass. Which is the number of protons and neutrons combined.

8 0

3 years ago

How many moles of each element (C, H, and O) are present in a 100.0 g sample of<br> ethanol?

Anuta_ua [19.1K]

Answer:

2.173 moles of ethanol is presented in a 100.0g sample of ethanol .

Explanation:

The amount of substance that contains as many Particles as there are atoms in exactly 12g of carbon- '12 isotope is called 1 mole '= 46 u.

7 0

3 years ago

Plz i need help please help

igomit [66]

Answer:

Metal

Explanation:

Metal is able to be shaped or stretched into wires without breaking

It is a good conductor because it has mobile electron and are solid in room temperature

Hope it helped :))

6 0

3 years ago

Read 2 more answers

Water heater contains 56 part a how many kilowatt-hours of energy are necessary to heat the water in the water heater by 25 ?c?

DochEvi [55]

Answer is: 550,021 kWh of energy is needed to heat the water

V(water) = 51 gal = 51·3,78 = 189,3 L.

ΔT(water) = 25°C.

d(water) = 1000 g/L.

m(water) = V(water) · d(water)

m(water) = 189,3 L · 1000 g/L

m(water) = 189300 g.

Q = m(water) · ΔT(water) · C(water)

Q = 189300 g · 25°C · 4,184 J/°C·g

Q = 19800780 J = 19800,78 kJ ÷ 3600 = 550,021 kWh.

6 0

3 years ago

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, $H_3PO_4$ acid and there are three Ka values

$K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}$

The transformation of $H_2PO_4- (aq) to HPO_4^2-(aq)$ is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

= 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

= 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

= -log 6.2x10-8

= 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

= 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0

3 years ago