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2 years ago

I have to do an experiment about reaction rates, and i need to have things to dissolve..

Chemistry

1 answer:

Leona [35]2 years ago

3 0

Answer:

salt ( with food coloring)

sand

sodium chloride

coffee

flour

oil

vinegar

different types of pancake mix

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How many moles are present in 1.75 g of zinc sulfate, Znso, ?

Cerrena [4.2K]

Explanation:

We assume you are converting between grams ZnSO4 and mole. You can view more details on each measurement unit: molecular weight of ZnSO4 or mol This compound is also known as Zinc Sulfate. The SI base unit for amount of substance is the mole. 1 grams ZnSO4 is equal to 0.0061941519772353 mole.

7 0

3 years ago

6. What causes the phases of the moon?

lawyer [7]

Answer:

The rotation of the Earth.

6 0

3 years ago

2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o phosphoric acid, h3po4, is neutralized by magnesium hydroxide, mg(oh)2, according to the eq

Rina8888 [55]

The balanced equation for the neutralisation reaction is as follows
2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O
stoichiometry of H₃PO₄ to H₂O is 2:6
number of H₃PO₄ moles reacted - 0.24 mol
if 2 mol of H₃PO₄ form 6 mol of H₂O
then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O
therefore 0.72 mol of H₂O are formed

4 0

3 years ago

Can somebody help me, please!

zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

Rubidium-85

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

72/100= 0.72 or 72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

Rubidium-87

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

28/100= 0.28 or 28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

Atomic Mass of Rubidium:

Add the two numbers together.

Rb-85 (61.2) and Rb-87 (24.36)

61.2+24.36=85.56 amu

4 0

3 years ago

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKa NH₃/NH₄⁺

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

Moles NH₃

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

H-H equation:

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

3 0

4 years ago