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Gre4nikov [31]
3 years ago
7

Tetraphosphorus decaiodide is represented by which formula? A) P3I10 B) P4I10 C) P3I8 D) P4I8

Chemistry
2 answers:
dusya [7]3 years ago
6 0

Answer: B) P_4I_{10}

Explanation: P_4I_{10} is a covalent compound because sharing of electrons takes place between phosphorous and iodine. Both the elements are non-metals and hence, will form covalent bond.

The nomenclature of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written then, and a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms , 'tetra' for 4 atoms and so on.

So, the IUPAC name for P_4I_{10} is tetraphosphorus decaiodide.

oksano4ka [1.4K]3 years ago
3 0
The answer is B.

Tetraphosphorus is written as P4, and decaiodide as I10, so when you put them together, it's P4I10
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Answer:

2192.64 PSI.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the container in L (V = 1650 L).

n is the no. of moles of the gas in mol (n = 9750 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas in (T = 35°C + 273 = 308 K).

∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.

  • <u><em>To convert from atm to PSI:</em></u>

1 atm = 14.696 PSI.

<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>

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3 years ago
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What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)
Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\  ml\\\\V = 33.33 \ ml

Therefore, volume required is 33.33 ml .

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In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to
labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of NH_3 = 100 g

Molar mass of NH_3 = 27 g/mole

Molar mass of N_2 = 28 g/mole

First we have to calculate moles of NH_3.

\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles

The given balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the given reaction, we conclude that

2 moles of NH_3 produced from 1 mole of N_2

3.7 moles of NH_3 produced from \frac{1mole}{2mole}\times 3.7mole=1.85moles of N_2

Now we have to calculate the mass of N_2.

Mass of N_2 = Moles of N_2 × Molar mass of N_2

Mass of N_2 = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0
3 years ago
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