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Lilit [14]
3 years ago
12

What is the charge on the calcium ion in Ca3(PO4)2?

Chemistry
1 answer:
irina1246 [14]3 years ago
5 0

Answer: calcium ion has +2 charge

Explanation:

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Which of the following is NOT a property of an acid?
Kryger [21]
The answer here is D - bases tend to feel slippery. This is because they tend to dissolve oils on your skin, reducing the friction between your fingers or hands when you rub them together.
7 0
3 years ago
A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro
Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)

So, OH^{-} is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of OH^{-} increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant.

Therefore excess amount of OH^{-} combines with NH_{4}^{+} to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
Consider the bonds in iron(III) oxide. Are the bonds more ionic or more covalent?
kvv77 [185]
<span>the bonds in iron(III) oxide are more ionic</span>
4 0
3 years ago
Read 2 more answers
How many water molecules are created when 32 grams of oxygen react completely with 4.0 grams of hydrogen?
ch4aika [34]

Answer:

d. 1.2 × 1024

Explanation:

From the equation of reaction

2H2 + O2= 2H2O

i.e 2mole(4g) of hydrogen requires 1 mole(32g) of oxygen to produce 2mole (2×6.02×10^23 molecules) of H2O= 1.2×20^24 molcules of water.

NB: 1 mole of H2O contains 6.02×10^23 molecules of H2O

3 0
3 years ago
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