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galben [10]
3 years ago
15

Assume you mixed 5 µl of your cod fish homogenate with 195 µl working solution. The fluorometer displays a measurement of 12.2 µ

g/ml. What is the protein concentration of the cod fish homogenate?
Chemistry
1 answer:
11111nata11111 [884]3 years ago
3 0

Answer:

Hallo

Explanation:

I took that question

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Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq
Rasek [7]

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s)  +  2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

7 0
3 years ago
Read 2 more answers
2. Balance the equation below, and answer the following question: What volume of chlorine gas, measured at STP, is needed to com
Juliette [100K]
Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)

when we have STP conditions, we can use this conversion: 1 mol = 22.4 L

first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation. 

molar mass of Na= 23.0 g/mol
 ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)

calculations:

6.25 g Na ( \frac{1 mol Na}{23.0 g} ) ( \frac{1 mol Cl_2}{2 mol Na} ) ( \frac{22.4 L}{1 mol} )= 3.04 L
8 0
3 years ago
The charge on an ion is known as its___number.
a_sh-v [17]

Answer:

oxidation number is correct!! :)

Explanation:

8 0
3 years ago
A student had a sample of pure water, and added an unknown substance to it. The student noticed that the hydroxide ion concentra
steposvetlana [31]
The answer is A it's a basic because once you add another substance to a neutral it either becomes acidic or basic. this one becomes basic because the hydroxide ion concentrate increased.
6 0
3 years ago
Read 2 more answers
Look at the image below:
PIT_PIT [208]

Answer:

The figure is a molecule and a compound

Explanation:

Because it have 2 cluster molecules

7 0
3 years ago
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