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zavuch27 [327]
3 years ago
5

Calculate the potential energy of a block of mass of mass 0.3kg and at a height 2.6m above the ground.(take acceleration due to

gravity:10m/s)​
Chemistry
2 answers:
algol [13]3 years ago
8 0
I don’t think I know this
Inga [223]3 years ago
5 0

Explanation:

Mass (m)=0.3 kg

height(h)=2.6 m

acceleration due to gravity (g)=10m/s^2

Potential energy= MGH

=0.3×10×2.6

7.8 J

Give brainliest if u can

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ycow [4]

Answer:

clouds???

Explanation:

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3 years ago
How do ions form bonds, and describe the structure of the resulting compound.​
Artemon [7]

Answer:

Electrons are transferred between atoms together in the ionic compound. The ions are arranged in a regular repeating pattern in an ionic crystal

Explanation:

6 0
2 years ago
I need three examples of objects with high density and three with low density!!
Anuta_ua [19.1K]

Answer:

For example, a suitcase jam-packed with clothes and souvenirs has a high density, while the same suitcase containing two pairs of underwear has low density. Size-wise, both suitcases look the same, but their density depends on the relationship between their mass and volume. Mass is the amount of matter in an object.

5 0
3 years ago
Read 2 more answers
Consider the following system at equilibrium:
alexgriva [62]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

6 0
3 years ago
You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
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