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kirza4 [7]
3 years ago
13

"a basic experiment involves a minimum of ________ participant group(s)."

Chemistry
1 answer:
3241004551 [841]3 years ago
5 0
Answer: TWO.

At least two groups: treatment group and control group.

The treatment group is that is exposed to the different levels of the independent variable ( a medication for example), while the control group is not treated, so the researchers can compare the effect of the medication.
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How many kilograms are in 6.983 moles of baking soda (NaCHO3)?
nasty-shy [4]
It should be about 0.586620881 kilograms
6 0
4 years ago
ILL GIVE YOU BRAINLIST !!! HAVE TO GET IT RIGHT !!
inysia [295]

Answer: The 3rd and 6th bullet point is the quantitative data.

Explanation: Quantitative data is expressed by NUMBERS and Qualitative data is expressed by WORDS. The 3rd and 6th one is correct because they both use numbers to compare how much time hummingbirds spent feeding on nectar.  

4 0
3 years ago
What is the bond order of li2−? express the bond order numerically?
jeka94
Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (ABMO). Bond Order is calculated as,

              Bond Order  =  (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2

              Bond Order  =  (4 - 3) ÷ 2

              Bond Order  =  1 ÷ 2
Or,
              Bond Order  =  1/2
Or,
              Bond Order  =  0.5

4 0
3 years ago
I don’t know what to say
Margaret [11]

Answer:

ugh it’s not coming out

Explanation:

5 0
2 years ago
​29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.
arsen [322]

The answer for the following mention bellow.

  • <u><em>Therefore the final temperature of the gas is 260 k</em></u>

Explanation:

Given:

Initial pressure (P_{1}) = 150.0 kPa

Final pressure (P_{2}) = 210.0 kPa

Initial volume (V_{1}) = 1.75 L

Final volume (V_{2}) = 1.30 L

Initial temperature (T_{1}) = -23°C = 250 k

To find:

Final temperature (T_{2})

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas  constant

T represents the temperature of the gas

We know;

\frac{P*V}{T} = constant

\frac{P_{1} }{P_{2} } × \frac{V_{1} }{V_{2} } = \frac{T_{1} }{T_{2} }

Where;

(P_{1}) represents the initial pressure of the gas

(P_{2}) represents the final pressure of the gas

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(T_{1}) represents the initial temperature of the gas

(T_{2}) represents the final temperature of the gas

So;

\frac{150 * 1.75}{210 * 1.30} = \frac{260}{T_{2} }

(T_{2}) =260 k

<u><em>Therefore the final temperature of the gas is 260 k</em></u>

<u><em></em></u>

3 0
3 years ago
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