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melamori03 [73]
2 years ago
6

What were the peak concentration for intravenous

Chemistry
1 answer:
Alenkinab [10]2 years ago
3 0
R1fhaszagvcnsjsjhacha
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Saturated aqueous sodium chloride(d=1.2g/mL) is added to the following mixtures in order to dry the organic layer. Which layer i
natali 33 [55]

(a) at the bottom - high density organic compound dissolved in methylene chloride

(b) at the bottom - saturated aqueous sodium chloride

Explanation:

Sodium chloride is dissolved in water while the organic compounds are dissolved in methylene chloride. After mixing the two solutions two layers will form because water (polar molecule) will not mix with the methylene chloride (nonpolar molecule).

The layer with higher density will be at the bottom.

(a) saturated aqueous sodium chloride (d = 1.2 g/mL) - upper layer

high density organic compound dissolved in methylene chloride (d = 1.4 g/mL) - bottom layer

(b) saturated aqueous sodium chloride (d = 1.2 g/mL) - bottom layer

low density organic compound dissolved in methylene chloride (d = 1.1 g/mL) - upper layer

Learn more about:

liquids with different densities

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7 0
3 years ago
What is the mass in grams of 2.50 mol of ammonia vapor, NH3?
Crank

Molar mass:-

\\ \tt\longmapsto 14+3(1)=17g/mol

Now

\\ \tt\longmapsto Given\:mass=No\;of\:moles\times Molar\;mass

\\ \tt\longmapsto Given\:mass=17(2.5)

\\ \tt\longmapsto Given\:mass=42.5g

7 0
2 years ago
When oxygen reacts with hydrogen, it has the capacity to release 29 kilojoules of energy. Inside a fuel cell, oxygen reacts with
Vanyuwa [196]

Answer:

79

Explanation:

7 0
3 years ago
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
Svet_ta [14]

Answer:

V_1=23.3~mL

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

C_1*V_1=C_2*V_2

Now we can identify the variables:

C_1=~1.475_M

V_1=~?

C_2=~0.1374_M

V_2=~250.0~mL

If we plug all the values into the equation:

1.475_M*V_1=0.1374_M*250.0~mL

And we solve for V_1:

V_1=\frac{0.1374_M*250.0~mL}{1.475_M}

V_1=23.3~mL

I hope it helps!

8 0
3 years ago
Convert 6.33×10−6 cg to nanograms.
tensa zangetsu [6.8K]
5.73e+8 hope this help
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3 years ago
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