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Over [174]
3 years ago
12

Which one of these elements is an alkaline earth metal

Chemistry
1 answer:
In-s [12.5K]3 years ago
7 0
The answer to this problem is Beryllium is an alkaline earth metal.
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How does the rock cycle and plate tectonics interfere with geochemists trying to determine the age of the Earth? Chose the best
Phoenix [80]

Answer:

The rock cycle and plate tectonics cause Earth's rocks to break down over time and they are recycled through natural processes.

Explanation:

Rock cycle(Attachment-1)

4 0
3 years ago
Which type of relationship exists between Alaskan brown bears and salmon? A. predation B. mutualism C. competition D. commensali
aksik [14]

Answer:

A. predation.

Explanation:

Predation is a type of ecological interaction in which one organism which is known as predator feeds upon another organism thereby killing it. The organism which is killed by predator is known as prey. In such interaction one organism is benefited while another one is harmed.

Brown bears of Alaska feed upon salmon. In order to so they have developed many ways which include waiting at the bottom of falls from where salmons pass by or standing at the bottom of the falls.

3 0
3 years ago
What was the 'elixir of life' ​
4vir4ik [10]
A magical or medicinal potion/solution
4 0
3 years ago
How many moles of AgCl2 are in 5.78x1024 molecules of AgCl2?
Zolol [24]

Answer:

9.6 mol AgCl2

Explanation:

You have to use Avogadro's number: 6.023 x 10^23

5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2

3 0
3 years ago
A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base
Gekata [30.6K]
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
                           = 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
                 = -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
      = 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
6 0
3 years ago
Read 2 more answers
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