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irakobra [83]
3 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]3 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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Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample
k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

<u>Step 1:</u> Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

<u>Step 2:</u> The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

<u>Step 3:</u> Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

<u>Step 4:</u> Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

<u>Step 5</u>: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0
3 years ago
A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *
elena-14-01-66 [18.8K]

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

  1. P1 = initial pressure (kPa)
  2. P2 = final pressure (kPa)
  3. V1 = initial volume (L)
  4. V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

3 0
3 years ago
Gasoline burning in an engine: Exothermic or Endothermic?
Alinara [238K]
An exothermic reaction releases heat. An endothermic reaction absorbs heat. Burning gas releases heat so it would be exothermic. Acid and water react heating the beaker would be exothermic because it releases heat from the reaction. Hope this helps! ;)
8 0
3 years ago
Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?
Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0
3 years ago
5. Determine the number of liters in 4.32 grams of water (H2O). Answer: --_ 5.37L H2O__​
MariettaO [177]

Answer:

4.32 \frac{1}{18.01} ×\frac{22.4}{1} = 5.37

Explanation:

22.4 is how many liters per mole. 18.01 is the mass of H2O.

8 0
3 years ago
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