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Slav-nsk [51]
3 years ago
7

The chemical or mechanical processes by which rocks when exposed to the weather undergo changes in character and break down are

called?
gravity sedimentation weathering magma
Chemistry
1 answer:
lukranit [14]3 years ago
7 0

Answer:

weathering

Explanation:

the rocks due to rain generate cracks . heat and wind helps the rock to break into smaller chunks .

this process continuous till the rock is converted into sand. and is called weathering

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How many grams of KO2 are needed to form 8.0 g of O2?
dem82 [27]
Using the atomic weights, K = 39, O=16 so KO2= 39+32=71. So K=39/71= 0.549 of KO2 and O2=32=0.45KO2 and O2=8g and 0.45KO2=8g so KO2=8/0.45=17.77g. So to check, K=0.549 x 17.77g=9.76 and so K+O2= 9.76+8g=17.76.
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4 years ago
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Which substance is an example of a colloid?
Semmy [17]

Answer:

Milk would be the only colloid

The sand and water, tomato juice and sugar and water are all known as suspensions (It consists of large particles mixed or suspended in a solution)

Explanation:

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3 years ago
Why does a salt compound give off light (or a colored flame) when burned? 4. did lithium chloride and sodium chloride give off s
Free_Kalibri [48]

3) Answer is: because elements have different emmision spectrum.

Emission spectrum of a chemical element is the spectrum of frequencies emitted due to an atom making a transition from a high energy state to a lower energy state. Each element's emission spectrum is unique.

4) Answer is: lithium chloride and sodium chloride do not have similar colors.

Lithium ion has red and sodium ion has yellow emission spectrum.

5) Answer is: because metals have valence electrons in different energy shells.

Each transition has a specific energy difference.

6) Answer is: salts of alkaline and earth alkaline metals.

For example barium chloride gives green light.

7 0
4 years ago
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BRAINLIESTTT !!! Pls help ASAP !!!! How does plasma state of matter interact with other matter?
Ivan

Answer:

Plasma is a state of matter in which an ionized gaseous substance becomes highly electrically conductive to the point that long-range electric and magnetic fields dominate the behaviour of the matter. The plasma state can be contrasted with the other states: solid, liquid, and gas

Explanation:

7 0
4 years ago
A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

  • E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ} and
  • m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}

Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

  • Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ} (converted to kilojoules)
  • Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}
  • Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

  • E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})
  • E(\text{released}) =  Q(\text{lead})

Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0
4 years ago
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