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Musya8 [376]
3 years ago
14

Logs in a fire, and charged batteries are examples of?

Chemistry
1 answer:
vlabodo [156]3 years ago
7 0

Answer:

Logs in a fire is an example of something burning, also changing its state of form. I think thats chemical change; going from wood to ashes. And charged batteries means they haven't been used yet or they were just charged, resulting in physical change.

Explanation:

I just learned this in science.

I hope this helps!

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What is the mass of 7.91 cm to the power of three piece of lead having a density of 11.34 g/cm to the power of three?
miss Akunina [59]

Hey there!

D = 11.34 g/cm³

V = 7.91 cm³

Therefore:

m = D * V

m = 11.34 * 7.91

m = 89.6994 g

4 0
3 years ago
If  O2(g) reacts with H2(g) to produce  H2O, what is the volume of H2O obtained from 1 L of O2?
joja [24]

Answer:

2L

Explanation:

We'll begin by writing the balanced equation for the reaction this is illustrated below:

2H2 + O2 —> 2H2O

1 mole of a gas occupy 22.4L.

1 mole of O2 occupy 22.4L.

2 moles of H2O occupy = 2 x 22.4 = 44.8L.

From the balanced equation above,

22.4L of O2 produced 44.8L of H2O.

Therefore, 1L of O2 will produce = 44.8/22.4 = 2L.

Therefore, 1L of O2 will produce 2L of H2O.

7 0
3 years ago
Describe how to use physical properties to seperate a mixture of ice cubes and nails
Marrrta [24]
The magnetic property of nails could be used to just pull them out with a magnet.
The difference in boiling points between water and iron or steel is hundreds of degrees Celsius so you could apply heat and just boil the water away.
4 0
3 years ago
Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s
disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid.  HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R   HSO_4^-    ⇄  H^+ + SO_4^2^-

I    0.14

C   - x               +x       +x

E   0.14-x        x         x

K_a = 1.3 × 10^-^2 for HSO^-_4 . As a result,

\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a

K_a is large. It is no longer valid to approximate that [HSO^-_4] at equilibrium is the same as its initial value.

\frac{x^2}{0.14-y} = 1.3 * 10^-^2

x^2+1.3*10^-^2x - 0.14 × 1.3 × 10^-^2= 0

Solving the quadratic equation for x , x \geq 0 since x represents a concentration;

                             x=0.0366538

Then, round the results to 2 significant figure;

  • [SO_4^2^-] = x = 0.037 mol. L ^-^1
  • [H^+] = x = 0.037 mol. L ^-^1
  • [HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1

Learn more about concentration here:

brainly.com/question/14469428

#SPJ4

3 0
1 year ago
DESPERATE WILL GIVE BRAINLIST AND THANKS
irina [24]

anaphase

or if not

telophase

6 0
3 years ago
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