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irakobra [83]
2 years ago
12

Give the chemical formula that corresponds to each of the following compounds: (a) Sodium tetrahydroxozincate(II) (b) Dichlorobi

s(ethylenediamine)cobalt(III) nitrate (c) Triaquabromoplatinum(II) chloride (d) Tetra-amminedinitroplatinum(IV) bromide
Chemistry
1 answer:
ladessa [460]2 years ago
5 0

Answer:

a. Na₂[Zn(OH)₄]

b. [Co(Cl)₂(en)₂]NO₃

c. [Pt(H₂O)₃Br]Cl

d. [Pt(NH₃)₃(NO₂)₂]Br₂

Explanation:

(a) Sodium tetrahydroxozincate(II)

Zincate(II) indicates the metal is Zn²⁺, tetrahydroxo are 4OH⁻, that is

[Zn(OH)₄]²⁻. As the molecule is a neutral salt of sodium:

<em>Na₂[Zn(OH)₄]</em>

(b) Dichlorobis(ethylenediamine)cobalt(III) nitrate

The metal is Co³⁺, bis(ethylenediamine) are (en)₂ and dichloro are 2 Cl⁻, that is:

[CoCl₂(en)]⁺

The counterion is NO₃⁻, that is:

<em>[CoCl₂(en)₂]NO₃</em>

(c) Triaquabromoplatinum(II) chloride

The metal is Pt²⁺, ligands are 3 H₂O and 1 Br⁻, that is:

[Pt(H₂O)₃Br]⁺

The counterion is Cl⁻:

<em>[Pt(H₂O)₃Br]Cl</em>

(d) Tetra-amminedinitroplatinum(IV) bromide

The metal is Pt⁴⁺, ligands are 2 NO₂⁻ anf 3 NH₃, that is:

[Pt(NH₃)₃(NO₂)₂]²⁺

As counterion is Br⁻ you need 2 Br⁻ for the neutral salt, that is:

<em>[Pt(NH₃)₃(NO₂)₂]Br₂</em>

<em></em>

I hope it helps!

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Which do you think would be safer for heating a flammable liquid sample contained in an open beaker, and Bunsen burner or an ele
Ray Of Light [21]

Answer:

Electric hot plate

Explanation:

Considering that the liquid sample is flammable, it wouldn’t be a good idea to have it near a source of fire which the Bunsen burner provides, rather it would be better to heat through an electric hot plate as its not near a source of flame and instead produces heat through electricity.

6 0
3 years ago
A zinc rod is placed in 0.1 MZnSO4 solution at 298 K. Write the electrode reaction and calculate the potential of the electrode.
yanalaym [24]

<u>Answer:</u> The potential of electrode is -0.79 V

<u>Explanation:</u>

When zinc is dipped in zinc sulfate solution, the electrode formed is Zn^{2+}(aq.)/Zn(s)

Reduction reaction follows:  Zn^{2+}(0.1M)+2e^-\rightarrow Zn(s);(E^o_{Zn^{2+}/Zn}=-0.76V)

To calculate the potential of electrode, we use the equation given by Nernst equation:

E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

[Zn]=1M    (concentration of pure solids are taken as 1)

[Zn^{2+}]=0.1M

Putting values in above equation, we get:

E_{cell}=-0.76-\frac{0.059}{2}\times \log(\frac{1}{0.1})\\\\E_{cell}=-0.79V

Hence, the potential of electrode is -0.79 V

4 0
3 years ago
Plz help I need to bring up my grade, this will help bring up my grade !!!!!!!
BartSMP [9]

Answer:

WHY: You can abbreviate an element's electron configuration using the noble gas notation method because when you get down to the lower elements, specifically the d's and the f's, the electron configuration will be very long. The noble gas notation method is a faster answer while also being correct.

HOW: We can abbreviate an element's electron configuration by finding the last noble gas a specific element passed, for example calcium would have just passed Argon. Once you have the "address" of the previous noble gas, then you add on the difference between the element chosen and the noble gas, for example calcium would be [Ar] 4s^2.

Explanation:

3 0
2 years ago
A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th
Ann [662]

Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

            Q = M x C x ΔT

            Q = -100 x 4.18 x (-5.14)

            Q = 2148.5 joules

            Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

          ΔH = Q / number of moles of NaHCO3

          ΔH = 2.1485 / 0.173

          ΔH = 12.42 KJ/mol

3 0
2 years ago
when 1 mol of glucose is burned, 2802.5kj of energy is released. calculate rhe quantity of energy released to a person by eating
tester [92]

77.78 kJ of energy is released when 1 mol of glucose is burned, 2802.5 kJ of energy is released.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 ×10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Calculate the moles of 5.00g of glucose.

Given mass = 5.00g

The molar mass of glucose = 180.156 g/mol

Moles = \frac{mass}{molar \;mass}

Moles = \frac{5.00g}{180.156 g/mol}

Moles =0.02775372455

The quantity of energy released to a person by eating 5.00g of glucose in a candy.

0.02775372455 x 2802.5 kJ

77.77981305 kJ =77.78 kJ

Hence, 777.78 kJ of energy is released.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

8 0
1 year ago
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