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drek231 [11]
4 years ago
5

I need help on this question please help

Chemistry
2 answers:
elixir [45]4 years ago
8 0
B. distance and time
liberstina [14]4 years ago
4 0
B. Distance and Time
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What material is least likely to be recognized as a mixture by looking under a microscope
I am Lyosha [343]
A homogenous mixture is uniform and thus hard to recognize as a mixture. An example is water.
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4 years ago
Compound a has molecular formula c5h10. hydroboration-oxidation of compound a produces 2-methylbutan-1-ol. draw the structure of
maw [93]
To help you answer this question, you can search a sample reaction in the internet for hydroboration-oxidation. Take a look at the attached picture. The sample reaction is shown at the top. We can deduce that the original compound must be an alkene. Following the pattern, compound a must be 2-methylbutene as shown in the picture.

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4 years ago
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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
4 years ago
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This leaf is green is A. an observation, B. a hypothesis. C. a theory As Corporation. All Rights Reseruar​
Trava [24]

Answer:

A. an observation

Explanation:

Because by saying that the leaf is green that is an observation on the leaf to tell what color it is.

Have a nice day/night :)

7 0
3 years ago
How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50
Aleks04 [339]

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

<h3>How to determine the pKa</h3>

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

  • Equilibrium constant (Ka) = 2.3×10¯⁵
  • pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

<h3>How to determine the molarity of HCl </h3>
  • pKa = 4.64
  • pH = 6.5
  • Molarity of salt [NaZ] = 0.5 M
  • Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

<h3>How to determine the mole of HCl </h3>
  • Molarity of HCl = 0.0069 M
  • Volume = 150 mL = 150 / 1000 = 0.15 L
  • Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

<h3>Complete question</h3>

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

Learn more about pH of buffer:

brainly.com/question/21881762

7 0
3 years ago
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