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Alex17521 [72]
3 years ago
6

Which best describes how sediment forms??

Chemistry
2 answers:
Andreas93 [3]3 years ago
8 0

Answer:

a.) the breakdown into small particles of rock and stuff

Firlakuza [10]3 years ago
6 0

Answer:

a

Explanation:

Hope this hel;ps!!

You might be interested in
5)
Drupady [299]

What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that

does its temperature vary by 25 ° C?

Answer:

143.75cal

Explanation:

Given parameters:

Mass of steel  = 50g

Specific heat capacity of the steel  = 0.115cal/g°C

Temperature  = 25°C

Unknown:

Amount of heat = ?

Solution:

The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.

  Amount of heat  = m C (ΔT)

m is the mass

c is the specific heat capacity

ΔT is the temperature change

 Now insert the parameters and solve;

  Amount of heat  = 50 x 0.115 x 25

  Amount of heat  = 143.75cal

4 0
2 years ago
Which sample is most likely to experience the smallest temperature change upon observing 55KJ of heat? 
Zigmanuir [339]

Answer:

100 g of water: specific heat of water 4.18 J/g°C

Explanation:

To know the correct answer to the question, we shall determine the temperature change in each case.

For 100 g of water:

Mass (M) = 100 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 4.18 x ΔT

Divide both side by 100 x 4.18

ΔT = 55000/ (100 x 4.18)

ΔT = 131.6 °C

Therefore the temperature change is 131.6 °C

For 50 g of water:

Mass (M) = 50 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 4.18 x ΔT

Divide both side by 50 x 4.18

ΔT = 55000/ (50 x 4.18)

ΔT = 263.2 °C

Therefore the temperature change is 263.2 °C

For 50 g of lead:

Mass (M) = 50 g

Specific heat capacity (C) = 0.128 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 0.128 x ΔT

Divide both side by 50 x 0.128

ΔT = 55000/ (50 x 0.128)

ΔT = 8593.8 °C

Therefore the temperature change is 8593.8 °C.

For 100 g of iron:

Mass (M) = 100 g

Specific heat capacity (C) = 0.449 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 0.449 x ΔT

Divide both side by 100 x 0.449

ΔT = 55000/ (100 x 0.449)

ΔT = 1224.9 °C

Therefore the temperature change is 1224.9 °C.

The table below gives the summary of the temperature change of each substance:

Mass >>> Substance >> Temp. Change

100 g >>> Water >>>>>> 131.6 °C

50 g >>>> Water >>>>>> 263.2 °C

50 g >>>> Lead >>>>>>> 8593.8 °C

100 g >>> Iron >>>>>>>> 1224.9 °C

From the table given above we can see that 100 g of water has the smallest temperature change.

5 0
3 years ago
What did Ernest Rutherford s gold foil experiment demonstrate about atoms?
mote1985 [20]
Rutherford performed gold foil experiment to understand that how negative and positive particles could Co exist in an atom. He bombarded alpha particles on a 0.00004 cm thick gold foil.

He proposed a planetary model of the atom and concluded following results and demonstrated that,
1. An atom produces a line spectrum.
2. An Electron revolves around the nucleus without any orbits.
3. Since most of the particles passed through the foil undeflected it means that most of the volume occupied by an atom is empty.
4. An Atom as a whole is neutral.
5. The deflection of few particles on the foil suggested that there is center of positive particles in an atom called the nucleus of the atom.
6. The complete rebounce of few particles on the gold foil suggested that the nucleus is very dense and hard.
6 0
3 years ago
Write the chemical formula for the cation present in the aqueous solution of cuso4.
Inessa05 [86]

The ionic compounds on dissolving in aqueous solution dissociate into their respective ions that are cations and anions.

The ions present in the aqueous solution of CuSO_4 are Cu^{2+} and SO_{4}^{2-}.

So, the chemical formula of the cation present in the aqueous solution of CuSO_4 is Cu^{2+}.

4 0
3 years ago
If I add water to 100ml of a 0.15m potassium sulfate until the final volume is 150ml what will the molarity of the diluted solut
Delicious77 [7]

Answer:

0.1 M is the answer and I have explained it in the attachment below.

Hope this helped!

8 0
3 years ago
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