y at 25 °C of Zn(OH), in pure water and in a 0.0140M ZnSO4 solution. You'll find K's, data sp nswers to 2 significant digits. ㅁ을 my ZnSO4 solution: 02 M ter: ㅁㅁ x10 *ㅁ ㅁㅁ x 5 ?

Q: 3. Solve the following problems, stating answers to the proper number of significant figures: a)…

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A:

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A:

Q: 1×10 -14 -8 = (6.4×10³ +x)x

A: Step 1: Given equation is:  1×10−14=(6.4×10−8+x)x x2+(6.4×10−8)x−10−14=0 ⇒x2+1086.4x​−10−141​=0 Tip:…

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A: The mass of gold is calculated in step two.

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A:

Q: 3.23x10^-23mg / (2.1x10^-10 cm)^3

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A: 1m = 103 mm  1m = 102 cm  1km = 103 m  1km = 105 cm 1 m3 = 109 mm3  1km2 = 1010 cm2  1 cm2 = 10-10…

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A: A question based on dimensional analysis that is to be accomplished.

Q: The correct answer to this was 1.8*10^-12 but how do you get to this Answer?

A:

Q: what is 1.0x10^-5 equal?

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Q: (5.0 × 104) × (3.0 × 106) =  a. 1.5 x 10^9 b. 1.5 x 10^10 c. 1.5 x 10^11 d. 1.5 x 10^12

A: The calculation given is, => (5.0 × 104 ) × (3.0 × 106)

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Q: (2.6*10^-6)*(2.02*10^4)

A: (2.6*10^-6)*(2.02*10^4) = =(2.6×10-6) × (2.02×104)=(2.6×2.02) × 10-6+4=5.25×10-2

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Q: Calculate the solubility at 25 °C of AgCl in pure water and in a 0.0090M AGNO, solution. You'll find…

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Q: 10^-2 + 9.92 x 10^-3

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Q: (7.211*10^5)*(5.3413*10^4)/4.09*10^-7

A:

Q: A student trying to determine the concentration of phosphate in an unknown solution produced the…

A: Given Absorbanve of unknown = 0.268

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Q: What is the mass of (1.000) Litre of gold? The density of gold is 19.3 grams/cm^3

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Q: From the equilibrium values Kb [4.79x 10-6+x] x [4.79 × 10-6+ X [0.8037-x] = 4.72 x 10-10 On solving…

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