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snow_lady [41]
2 years ago
9

A 1.19 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen

Chemistry
1 answer:
Dimas [21]2 years ago
7 0

Answer:

The halogen is Iodine.

Explanation:

Using the ideal gas equation, we find the number of moles of gas present, n.

PV = nRT where P = pressure of gas = 1.41 atm, V = volume of gas = 109 mL = 0.109 L, n = number of moles of gas, R = molar gas constant = 0.082 L-atm/mol-K and T = temperature of gas = 398 K

Since PV = nRT, making n subject of the formula, we have

n = PV/RT

substituting the values of the variables into the equation, we have

n = 1.41 atm × 0.109 L/(0.082 L-atm/mol-K × 398 K)

n = 0.15369 atm-L/32.636 L-atm/mol

n = 0.0047 mol

Since n = m/M where m = mass of gas = 1.19 g and M = relative molecular mass of gas

So, M = m/n

M = 1.19 g/0.0047 mol

M = 252.7 g

Since halogens are diatomic the relative atomic mass is M/2 = 252.7g/2 = 126.34 g

From tables, the only halogen with this atomic mass is Iodine.

So, the halogen is Iodine.

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Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

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

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

<u>Step 2:</u> Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

Q = 50208 Joule = 50.2 kJ

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8 0
2 years ago
Which of the Atoms shown has an atomic number four
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4 0
2 years ago
A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added
kiruha [24]
<h3><u>Answer;</u></h3>

pH = 12.33

<h3><u>Explanation;</u></h3>

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

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Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.  

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

7 0
3 years ago
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Answer:

\boxed{It\ will\ shift\ to\ create\ less\ of\ substance\ A}

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If the concentration of any substance A in a dynamic equilibrium increases, The equilibrium will be  shifted to its opposite side so that Substance A can be created less and the substance opposite to A can be created more so that a "dynamic equilibrium" can again be established.

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