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sdas [7]
3 years ago
9

_____ are the results of a thoroughly tested hypothesis?

Chemistry
1 answer:
Andru [333]3 years ago
6 0

i think its theory but im not to sure

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2 years ago
Brownian motion is not observed in a mixture that includes two phases. Which best identifies the mixture? [A} a colloid
Hoochie [10]

b, a suspension

Hope this helped :)

3 0
3 years ago
Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final
Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

<u>Step 1:</u> Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

<u>Step 2:</u> The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

<u>Step 3:</u> Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

<u>Step 4: </u>Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

<u>Step 5:</u> Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

<u>Step 6:</u> Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0
3 years ago
What is the molar mass of Al^2(SO^4)3
borishaifa [10]
342.14 g/mol

Molar mass of Al= 26.98
Molar mass of S=32.06
Molar mass of O=16.00

(26.98)2+(32.06+(16.00×4))3=342.14
7 0
3 years ago
Can anyone help me with this please​
ddd [48]

Answer:

5 = particle 7= atoms

Explanation:

i renever easily

4 0
3 years ago
Read 2 more answers
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