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Klio2033 [76]
3 years ago
14

How many elements occur naturally on earth?

Chemistry
1 answer:
Sati [7]3 years ago
6 0
There are 92 that appear on the periodic table. However, there are elements that occur in trace amounts naturally which brings the number to 98 natural elements on Earth.
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Which family of elements can be found in the area marked B
postnew [5]

Answer:

A) Alkaline Earth Metals

Explanation:

Area Marked A would be Alkali Metals

Area Marked B would be Alkaline Earth Metals

Area Marked C would be Transition Metals

Area marked H would be Halogens

3 0
3 years ago
How are plate tectonics related to earth's surface features?
melomori [17]
It is because due to the movement of the plates, natural diasters such as earthquakes take place on the earths surface.

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5 0
4 years ago
How can we increase the rate of collisions between the reactants in this reaction?
Juli2301 [7.4K]

Answer:

According to the collision theory, we can increase the rate of a reaction by increasing the concentration of reactants. We can also increase the rate of a reaction by increasing temperature to safe levels.

In the reaction mentioned in the question, Mg + 2HCl → MgCl2 + H2, if we increase the amount of magnesium or HCl, the reaction will proceed faster. Similarly, if we increase the temperature, then also the reaction rate will increase.

8 0
3 years ago
Read 2 more answers
Which of the following pairs would make a good buffer? A. H2C2O4 and HC2O4- B. HCl and NaOH C. CH3COOH and NaOH D. HNO3 and NO3-
pogonyaev
The answer is A. H2C2O4 and HC2O4-. For B and C, the two substances will reaction with each other. For A, it can be protons donor and receptor at the same time. So it is a good buffer.
5 0
4 years ago
What is the specific heat of a 44 g piece of metal if 202 Joules of heat are required to
Ray Of Light [21]

Answer:

≈ 395,8 J/(kg * °C)

Explanation:

m = 44 g = 0,044 kg

t_{1} = 22 °C

t_{2} = 33,6 °C

Q = 202 J

The formula is: Q = c * m * (t_{2} - t_{1})

c = \frac{Q}{m * (t_{2} - t_{1} )}

Calculating:

c = 202 J / 0,044 kg * (33,6 °C - 22 °C) ≈ 395,8 J/(kg * °C)

4 0
3 years ago
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