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Nadusha1986 [10]
3 years ago
6

THE WEATHER BALLOON DATA WE STUDIED WAS MEASURED AND RECORDED AT MIDNIGHT. THE TEMPERATURE OF THE AIR AT Earth's SURFACE IS DIFF

ERENT DURING DIFFERENT TIME OF THE DAY. WHAT OTHER TIMES OF THE DAY COULD WE MEASURE THE SURFACE TEMPERATURES OUTSIDE THE SCHOOL BUILDING? WHY? Write your thought on a sticky note and add it to our board. (Please initial your note.)​
Chemistry
1 answer:
melomori [17]3 years ago
3 0

Answer night

Explanation:

because day is to hot

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A 15.0 mL sample of a liquid has a mass of 12.3 g. What is the density of the liquid? A. 0.820g/mL B. 1.22g/mL C. 185g/mL D. 12.
MariettaO [177]

Answer: Option A. 0.820g/mL

Explanation:

Mass = 12.3 g

Volume = 15mL

Density = Mass /volume

Density = 12.3/15 = 0.82g/mL

4 0
3 years ago
Consider pure water separated from an aqueous sugar solution by a semipermeable membrane, which allows water to pass freely but
mojhsa [17]

Answer:

it will be the same on both sides of the membrane

7 0
3 years ago
a phase change from B to A will represent a loss of energy. what is the term to describe this transition if B is the liquid phas
Reptile [31]

Answer:

freezing

Explanation:

The process in which a liquid changes to a solid is called freezing or solidification. Energy is taken away during freezing.

5 0
4 years ago
How many significant figures are in 0.07080 g?
goldenfox [79]
There are 4 significant figures. 
zeros after the decimal point are not significant
7 0
3 years ago
For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ
Komok [63]

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

6 0
3 years ago
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