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Stella [2.4K]
3 years ago
12

How are land air and water heated each day?

Chemistry
1 answer:
almond37 [142]3 years ago
4 0
They are heated by the sun. Good luck :)
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HELP ASAP 50-50 CHANCE
docker41 [41]

Answer: The answer is A, A new element or different atom formed from the original two.

Hope this helps! :D

-<em>TanqR</em>

4 0
3 years ago
Accuracy vs. Precision
Oksi-84 [34.3K]

Answer:

Accuracy is the closeness to the specific target and precision is the closeness of the measurements to each other.

3 0
3 years ago
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Help Plz 20 points -------------
Arturiano [62]

Answer:

B

Explanation:

Compounds are made up of 2 or more different atoms that bond to form a compound.

4 0
2 years ago
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A chemist adds 0.60L of a 0.20/molL sodium thiosulfate Na2S2O3 solution to a reaction flask. Calculate the millimoles of sodium
Serggg [28]

Answer:

1.2×10² mmole of Na₂S₂O₃

Explanation:

From the question given above, the following data were obtained:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:

Volume = 0.6 L

Molarity = 0.2 mol/L

Mole of Na₂S₂O₃ =?

Molarity = mole /Volume

0.2 = Mole of Na₂S₂O₃ / 0.6

Cross multiply

Mole of Na₂S₂O₃ = 0.2 × 0.6

Mole of Na₂S₂O₃ = 0.12 mole

Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:

1 mole = 1000 mmol

Therefore,

0.12 mole = 0.12 mole × 1000 mmol / 1 mole

0.12 mole = 120 = 1.2×10² mmole

Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃

7 0
3 years ago
calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate
kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

#SPJ4

4 0
1 year ago
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