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nlexa [21]
3 years ago
12

What is a ‘control’ in an experiment?

Chemistry
2 answers:
Over [174]3 years ago
7 0

D. The name for the set of independent and dependent variables that will be controlled by the scientist.

tensa zangetsu [6.8K]3 years ago
3 0

Answer:

The answer fo this is D because of the person controlling the outcome for both variables

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If a patient is injected with 0.500 L of IV saline, what is the volume in quarts? (Given: 1 qt = 946 mL)
Andrei [34K]

Answer:

0.529 qts

Explanation:

5 0
3 years ago
Read 2 more answers
Determine the molar mass of carbon dioxide (CO2)
Sergio039 [100]
<span>44.01 g/mol hope it helps</span>
3 0
3 years ago
a balloon contains 30.0 L of helium gas at 103 kPA. what is the volume of the helium when the balloon rises to an altitude where
stiv31 [10]
Answer:
                V₂  =  123.6 L

Explanation:

According to Boyle's law pressure and volume of a gas are inversely related if amount and temperature are kept constant. For the initial and final states the gas law is given as,

                                                P₁ V₁  =  P₂ V₂     ----- (1)
Data Given;
                   P₁  =  103 kPa

                   V₁  =  30 L

                   P₂  =  25 kPa

                   V₂  =  ?

Solution:

Solving equation 1 for V₂,

                                        V₂  =  P₁ V₁ / P₂

Putting values,
                                        V₂  =  (103 kPa × 30 L) ÷ 25 kPa

                                        V₂  =  123.6 L

Result:
           As the pressure is decreased from 103 kPa to 25 kPa, therefore, volume has increased from 30 L to 123.6 L.
8 0
3 years ago
1.
tester [92]

Answer: New moon!

Explanation: So if your saying lunar eclips it would be new moon becuase we have full moons all the time that means lunar eclipses all the time soo it would be new moon

5 0
3 years ago
Read 2 more answers
A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that fo
amid [387]

Explanation:

Ksp of NiCO3 = 1.4 x 10^-7

Ksp of CuCO3 = 2.5 x 10^-10

Ionic equations:

NiCO3 --> Ni2+ + CO3^2-

CuCO3 --> Cu2+ + CO3^2-

[Cu2+][CO3^2-]/[Ni2+][CO3^2-]

= (2.5* 10^-10)/(1.4* 10^-7)

= 0.00179.

[Cu2+]/[Ni2+]

= 0.00179

= 0.00179*[Ni2+]

If all of Cu2+ is precipitated before Na2CO3 is added.

= 0.00179 * (0.25)

The amount of Cu2+ not precipitated = 0.000448 M

The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100

= 0.000448/0.25 * 100

= 0.18%

Therefore, percentage precipitated = 100 - 0.18

= 99.8%

The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.

8 0
3 years ago
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