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2 years ago

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During the analysis, 0.163 g H2O and 0.600 g CO2 are produced. Calcuate the amount (mol) C in the sample, considering the mass o

f CO2 gas formed by combustion.
0.00905 mol H2O formed

0.0181 mol H in sample

2 answers:

SSSSS [86.1K]2 years ago

6 0

Answer:

The amount of carbon is the sample is 0.163 grams.

Explanation:

Mass of carbon dioxide produced by the sample = 0.600 g

Moles of carbon dioxide = $\frac{0.600 g}{44 g/mol}=0.0136 mol$

Moles of carbon atom in 0.01364 moles of carbon dioxide:

1 × 0.0136 mol = 0.0136 mol

Mass of 0.01364 moles of carbon :

$=0.0136 mol\times 12 g/mol=0.163 g$

The amount of carbon is the sample is 0.163 grams.

galina1969 [7]2 years ago

5 0

Answer:

0.0136 mol

Explanation:

C ⟶ CO₂

1. Moles of CO₂

$\text{Moles of CO}_{2} = \text{0.600 g CO}_{2} \times \dfrac{\text{1 mol CO}_{2}}{\text{44.01 g CO}_{2}} = \text{0.013 63 mol CO}_{2}$

2. Moles of C

$\text{Moles of C} = \text{0.013 63 mol CO}_{2} \times \dfrac{\text{1 mol C}}{\text{1 mol CO}_{2}} = \textbf{0.0136 mol C}$

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A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl

ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

$AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}{diluted}]$ = 0.0222 M

$[Cl^{-}{concentrated}]$ = 2.22 M

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$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V

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2 years ago

In the reaction 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g), what compound is in the gaseous state?

ipn [44]

Answer is: C. H₂, molecule of hydrogen, g is c<span>hemistry abbreviations or physical state symbol for gas.</span>
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Lithium (Li) is solid (s) element (metal).
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2 years ago

HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

joja [24]

Answer:

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Explanation:

Hello,

In this case, for the given dissociation, we have the following equilibrium expression in terms of the law of mass action:

$Ka=\frac{[H_3O^+][BrO^-]}{[HBrO]}$

Of course, water is excluded as it is liquid and the concentration of aqueous species should be considered only. In such a way, in terms of the change $x$ , we rewrite the expression considering an ICE table and the initial concentration of HBrO that is 0.749 M:

$5.2x10^{-5}=\frac{x*x}{0.749-x}$

Thus, we obtain a quadratic equation whose solution is:

$x_1=-0.00627M\\x_2=0.00624M$

Clearly, the solution is 0.00624 M as no negative concentrations are allowed, so the concentration of BrO⁻ is 6.24 x 10-3 M.

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2 years ago

Read 2 more answers

A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

zhannawk [14.2K]

Answer: The molecular formula will be $C_8H_8O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

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2 years ago

An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

<u>Answer:</u> The unknown salt is NaF

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

<u>Initial:</u> 0.1

<u>At eqllm:</u> 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

2 years ago