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horsena [70]
3 years ago
12

What is a control group and is it always needed in an experiment?

Chemistry
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer:

yes, it is always needed

Explanation:

a control group is what you choose to control in an expirement its the amount of somthing you don't change

Hope this helped

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How is the food in a stomach broken down into simpler substances? What chemicals help this process?
Ksenya-84 [330]

Answer:

with the help of the juice contained in it

4 0
3 years ago
How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that
tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,  

= 2.50 L × 0.800 mol/L

= 2 mol

The given pH of a buffer is 8.53

pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,  

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:  

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,  

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.  

8 0
3 years ago
Both AM and FM radio signals have the same of what.....
sweet [91]
Both AM and FM radio signals have the same frequencies at times.
3 0
3 years ago
A chemist lectures about the following definition for acids and bases. The acid-base behavior is analyzed in terms of how electr
Y_Kistochka [10]

Answer:

the Lewis concept of acids and bases

Explanation:

8 0
3 years ago
Read 2 more answers
How much heat is required to increase the temperature of 100.0 grams of iron from 15.0oC to 40.2oC? (The specific heat of iron i
Ksenya-84 [330]
Answer is: 1160 J of heat Is required to increase the temperature.
m(Fe) = 100 g.
∆T = 40,2 - 15 = 25,2°C.
C(Fe) = 0,46 J/g•°C.
Q = m(Fe) • C • ∆T.
Q = 100 g • 0,46 J/g•°C • 25,2°C
Q = 1160 J.
C - specific heat.

3 0
3 years ago
Read 2 more answers
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