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pogonyaev
3 years ago
11

Which of the following choices represents elements with the most similar

Chemistry
1 answer:
matrenka [14]3 years ago
5 0

Answer:

Explanation:

F Cl Br belongs to the Same group

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If the Earth were to spin half as fast as it does now, how many hours would be in one day (one complete spin)?
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Answer:

12hrs

Explanation:

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4 0
3 years ago
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Which three temperature readings all mean the same thing?
allochka39001 [22]

Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.

Explanation:

Conversion of degree Celsius to Kelvin :

K=^oC+273

Conversion of degree Celsius to degrees Fahrenheit :

^oF=(\frac{9}{5}\times ^oC)+32

By using these two conversion factors, we get the three temperature readings all mean the same thing.

For option A :

K=^oC+273=100+273=373K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF

For option B :

K=^oC+273=100+273=373K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF

For option C :

K=^oC+273=0+273=273K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF

For option D :

K=^oC+273=0+273=273K

^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF

From the given options, only option (D) is correct.

Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit

Hope this helps!

8 0
3 years ago
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What is the molality of an aqueous solution that contains 29.5 g of glucose (C6H12O6) dissolved in 950 g of water (H2O)?
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Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

Molality = 29.5 g glucose (1 mol / 180.16 g ) / .950 kg water
Molality = 0.1724 mol / kg
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If a buffer solution is 0.130 M in a weak acid (K_a = 1.7 x 10^-5) and 0.590 M in its conjugate base, what is the pH?
Serggg [28]
Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43
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