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larisa86 [58]
3 years ago
13

What are minerals in the rest of the water.

Chemistry
1 answer:
garri49 [273]3 years ago
5 0

Answer:

-calcium

-magnesium.

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What is the pOH of a<br> 5.6 x 10-5 M solution of cesium<br> hydroxide (CsOH)?
Rudiy27

Answer: 4.25

Explanation: CsOH=Cs+

pOH= -log10

= -log10(5.6*10^-5)

= -log10(5.6*10^-5)=-(-4.25)

=4.25

7 0
3 years ago
Which of the following best explains why bonds form between atoms?
ANEK [815]

Answer:

D

Explanation:

I believe the answer is D because atoms are always seeking to fill up their outer electron shell/valence shell and want to gain a full octet.

6 0
2 years ago
Identify two factors that can produce metamorphic rocks
alexdok [17]
Heat & pressure. hope this helps
8 0
3 years ago
Read 2 more answers
What is the correct answer ?
AlexFokin [52]

Answer:

Element with 6s subshell

Explanation:

Reactivity of an element depends on the electronic configuration and position of element in the periodic table as reactivity increases as we go down the periodic table.

This is so because number of shell increases as move down the periodic table and the last electron is further away from the nucleus.

Element with 6s subshell is the largest among 3s and 4s subshell and has more number of shells so it will react more than 3s and 4s subshell.

Hence, the correct answer is "Element with 6s subshell".

3 0
3 years ago
In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2
allsm [11]

Answer:

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

Explanation:

SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)

Balance the charge by adding 2 electrons on product side:

Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-....[1]

Reduction :

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)

Balance O by adding water on required side:

SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

Now, balance H by adding H^+ on the required side:

SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)

At last balance the charge by adding electrons on the side where positive charge is more:

SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)..[2]

Adding [1] and [2]:

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

The final balanced equation is :

SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)

4 0
2 years ago
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