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Gnom [1K]
3 years ago
5

(a) Give systematic names or the chemical formulas for the following. Remember, names must

Chemistry
1 answer:
lions [1.4K]3 years ago
7 0

Answer:

[Ni(CO)2C2O4]- Dicarbonyloxalatonickel(0)

K3[Fe(NCO)3(OCN)3-potassiumtricyanatotriisocyanatoferrate II

diammineaquahydroxoplatinum(II) ion- [Pt(NH3)2(H2O)(OH)]^+

Al[Co(CO3)2(CN)2- Aluminiumdicarbonatodicyanocobalt III

ethylenediaminedinitritozinc(II)- Zn(en)(NO2)2

Explanation:

The international Union of Pure and applied chemistry published a set rules for the nomenclature of inorganic compounds in 1985. This set of rules have been consistently revised ever since.

The rules stipulate the order for naming positive ions, negative ions and neutral ligands and the allowed order of preference for naming. It also stipulates the proper endings for ligands which are positive ions, negative ions or neutral ligands as the case may be.

The compounds named above were named in accordance with the revised IUPAC nomenclature for inorganic compounds.

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True or False: As the ice warms up, the molecules get more energy and move around more from place to place to form liquid water.
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The answer is TRUE.

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3 years ago
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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
Svetllana [295]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

7 0
4 years ago
Some kayakers coat the bottom of their kayaks with wax to make them more slippery. Kayakers wax their boats to
Sindrei [870]
The answer is : C. reduce their input force
they put wax on their kayak to make their boat cut smoothly through the water, which make it able to obtain further distance with less input force, which will increase it's overall speed 
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3 years ago
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Considering particles at the subatomic level, carrying out this experiment would help to identify the metals given that: Ca has
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Answer:

we can do it again and again and again and again and again and again

4 0
3 years ago
Estimate your de Broglie wavelength when you are running. (For this problem use h = 10^−34 in SI units and 1 lb is equivalent to
AlekseyPX

Answer:

A. your running speed 1.5 m/s

B. your mass 70 kg

C. your de Broglie wavelength 6.32x10^{-36}m

Explanation:

Hello there!

In this case, since the equation for the calculation of the Broglie wavelength is:

\lambda =\frac{h}{m*v}

We can assume a running speed of about 1.5 m/s and a mass of 70 kg, so the resulting Broglie wavelength is:

\lambda =\frac{6.626x10^{-34}kg\frac{m}{s} }{70kg*1.5\frac{m}{s} }\\\\\lambda =6.32x10^-36m

Best regards!

7 0
3 years ago
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