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trapecia [35]
1 year ago
10

Does any solid Ag₂CrO₄ form when 2.7x10⁻⁵g of AgNO₃ is dissolved in 15.0 mL of 4.0x10⁻⁴MK₂CrO₄?

Chemistry
1 answer:
e-lub [12.9K]1 year ago
3 0

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

Balanced reaction-:

<h3>2AgNO3(aq)+K2CrO4(aq)→Ag2CrO4(s)+2KNO3(aq)</h3>

Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol

=1.589⋅10^−7 mol

Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M

Ksp of Ag2CrO4

=[Ag+]2[CrO42−]

1.2⋅10−12=[2s]2[s]

4s3=1.2⋅10−12

s=6.69⋅10−5 M

Molarity of Ag+ is less than the molar solubility thus ppt will not occur.

<h3>What is the molarity calculation formula?</h3>

The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.

Learn more about Molarity:

brainly.com/question/8732513

#SPJ4

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Explanation:

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If liquid water is exposed to normal atmospheric pressure, what needs to change in order to change its state of matter?
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three students made multiple weighings of a copper cylinder each using a different balance. Describe the accuracy and precision
ch4aika [34]
I think that you have put up an incomplete question. However, i am answering the question based on my research and knowledge. 

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3 0
2 years ago
PLEASE HELP!!!!!!!!!
andreev551 [17]

The average atomic mass of the imaginary element : 47.255 amu

<h3>Further explanation  </h3>

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

Mass atom X = mass isotope 1 . % + mass isotope 2.% ..

isotope E-47 47.011 amu, 87.34%

isotope E-48 48.008 amu, 6.895

isotope E-49 50.009 amu, 5.77%

The average atomic mass :

\tt avg~mass=0.8734\times 47.011+0.06895\times 48.008+0.0577\times 50.009\\\\avg~mass=41.059+3.310+2.886\\\\avg~mass=47.255~amu

5 0
2 years ago
A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me
DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0
3 years ago
Read 2 more answers
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