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Masteriza [31]
3 years ago
14

Please help me, I am struggling please help

Chemistry
1 answer:
saw5 [17]3 years ago
8 0
Answer:

Colors are seen by burning of metal salts, for example Sodium chloride makes yellow frame.
It is very common for fireworks to contain Aluminum, Iron, Steal, Zinc or the Magnesium’s dust to create bright light.
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How many molecules of CaCi2 are equivalent to 75.9 grams
horsena [70]
<em>M CaCl₂: 40+(35,5×2) = 111 g/mol</em>


6,02·10²³ molecules ---------- 111g
X molecules --------------------- 75,9g
X = (75,9×<span>6,02·10²³)/111
X = <u>4,116</u></span><span><u>·10²³</u> molecules of CaCl</span>₂

:)
8 0
3 years ago
Which of the following statements is generally TRUE?
Vesnalui [34]

The correct answer is option d, that is, the solubility of a solid is highly dependent on temperature.  

Solubility refers to the maximum amount of a component, which will get dissolved in a given concentration of solvent at a particular temperature. The temperature influences the solubility of both gases and solids. The temperature has a direct influence on solubility.  

For most of the ionic solids, enhancing the temperature elevates how briskly the solution can be formed. With the increase in temperature, the movement of the solid particles takes place briskly that enhances the chances that they will associate with the majority of the solvent particles. This leads to enhancing the rate at which the solution takes place.  


5 0
3 years ago
When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --&gt; 2 AlCl3 (s) c) How many moles of the e
sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂  

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

                2 Al(s)  +  3Cl₂(g)  ⇒   2AlCl₃

Process

a) Calculate the molar mass of the reactants  

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion  Al/Cl₂ = 53.96/212.7 = 0.254

   Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

                    53.96 g of Al ------------------ 212.7 g of Cl₂

                     40.5 g of Al -------------------  x

                      x = (40.5 x 212.7) / 53.96

                      x = 8614.35 / 53.96

                      x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

                  = 53.057 g

d) Calculate the moles of Cl

                       35.45 g of Cl ----------------- 1 mol

                       53.057 g of Cl ---------------  x

                        x = (53.057 x 1)/35.45

                       x = 1.49 mol of Cl₂                    

6 0
3 years ago
Read 2 more answers
Question
hram777 [196]
I think it’s false : D
4 0
3 years ago
ASTRONOMY HELP! (:
Novay_Z [31]
C because it’s talking about the earth and it’s rotating
7 0
3 years ago
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