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aliya0001 [1]
3 years ago
7

Imagine you are on a roller coaster. Imagine that you are riding a skateboard or running across a ball field. What are some clue

s that you are moving?
Chemistry
1 answer:
Alex787 [66]3 years ago
4 0

✯Hello✯

↪  You can see the setting changing around you

↪  You can feel a higher pressure/wind on your face

↪  There is adrenaline going through you

↪  (if this isnt the context you are looking for just comment)

❤Gianna❤

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5) A scientist adds 1.2 liters of water to a 5.0 liter solution, causing the molarity to become 0.4 M. What was the molarity of
ivanzaharov [21]

Explanation:

i hope i helped you in this problem

5 0
3 years ago
Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.
Inga [223]

Answer:

1.5g/cm³

Explanation:

density=mass÷volume

mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g

volume of the cube = 10×10×10 = 1000cm³

density= divide 1500g÷1000cm = 1.5g/cm³

<h2>Density= 1.5g/cm³</h2>

YOUR WELCOME!

4 0
3 years ago
What is the general trend in electro negativity as you move from left to right across the periodic table
viva [34]

Answer:

On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you go down a group. As a result, the most electronegative elements are found on the top right of the periodic table, while the least electronegative elements are found on the bottom left.

Explanation:

6 0
3 years ago
Which question can be answered using the scientific method ?
anyanavicka [17]
The correct answer is D because the the rest will have different answers depending on the person you ask. D can be proven with facts by research
8 0
3 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
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