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sergij07 [2.7K]
3 years ago
7

Chemical formula 3CuCl2 + 2Al --> 2AlCl3 + 3Cu

Chemistry
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

nnkkkkhjkkhuokjhllkllllllll

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A box of jello has a mass of 250 g. How many boxes must be bought to have 1 Kg of jello? *
zhuklara [117]

Answer:

Number of boxes = 4

Explanation:

Given:

Mass of one box of jello = 250 grams

Total quantity want to purchase = 1 kg = 1 × 1,000 gram = 1,000 grams

Find:

Number of boxes in 1,000 grams = ?

Computation:

Number of boxes = Total quantity want to purchase / Mass of one box of jello

Number of boxes = 1,000 / 250

Number of boxes = 4

Therefore, 4 boxes of jello must be purchase to get 1 kg of Jello.

8 0
3 years ago
Uranium (VIII) Sulfide formula
tresset_1 [31]

Answer:

US₂

Explanation:

Uranium sulfide (US₂)

Uranium atomic symbol = U

Sulfur atomic symbol = S

Uranium valency = +4

Sulfur valency = -2

So;

Uranium sulfide (US₂)

8 0
3 years ago
how many grams of potassium chloride, KCL(molar mass is 74.55g/mol) are produced if 25 grams of potassium chlorate KClo^3 decomp
JulsSmile [24]

Answer:

                     Mass = 15.20 g of KCl

Explanation:

                    The balance chemical equation for the decomposition of KClO₃ is as follow;

                                            2 KClO₃ = 2 KCl + 3 O₂

Step 1: Calculate moles of KClO₃ as;

Moles = Mass / M/Mass

Moles = 25.0 g / 122.55 g/mol

Moles = 0.204 moles

Step 2: Find moles of KCl as;

According to equation,

                  2 moles of KClO₃ produces  =  2 moles of KCl

So,

               0.204 moles of KClO₃ will produce  =  X moles of KCl

Solving for X,

                       X = 2 mol × 0.204 mol / 2 mol

                       X = 0.204 mol of KCl

Step 3: Calculate mass of KCl as,

Mass = Moles × M.Mass

Mass = 0.204 mol × 74.55 g/mol

Mass = 15.20 g of KCl

6 0
3 years ago
Anyone want to talk ??​
3241004551 [841]
No is the answer your welcome well I don’t.
7 0
2 years ago
2. A sample of oxygen gas is placed in a rigid 1.5L glass container at STP. If the gas is
sladkih [1.3K]

The new pressure : P₂ = 1038.39 mmHg

<h3>Further explanation</h3>

Given

1.5 L container at STP

Heated to 100 °C

Required

The new pressure

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).

So P₁ =  1 atm = 760 mmHg

T₁ = 273 K

T₂ = 100 °C+273 = 373 K

Gay Lussac's Law  

When the volume is not changed, the gas pressure is proportional to its absolute temperature  

\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

Input the value :

P₂=(P₁.T₂)/T₁

P₂=(760 x 373)/273

P₂ = 1038.39 mmHg

7 0
3 years ago
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