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Sauron [17]
1 year ago
11

How many grams of carbon are contained in one mole of

e="C_{3}H_{8}" alt="C_{3}H_{8}" align="absmiddle" class="latex-formula"> ?
Chemistry
1 answer:
NNADVOKAT [17]1 year ago
4 0

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:36 \:\: g

____________________________________

\large \tt Explanation \: :

The given compound has 3 carbon atoms, so in 1 mole of that compound, there will be 3 moles of carbon atoms.

Mass of each mole of carbon atoms = 12 g

For 3 mole carbon atoms, it will be 12 × 3 = 36 grams

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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6 0
2 years ago
Let us assume that fe(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented
Yuki888 [10]

17.8 mL NaOH

<em>Step 1.</em> Write the chemical equation

Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)

<em>Step 2.</em> Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]

= 11.50 mmol Fe^(2+)

<em>Step 3.</em> Calculate the moles of NaOH

Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]

= 23.00 mmol NaOH

<em>Step 4.</em> Calculate the volume of NaOH

Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)

= 17.8 mL NaOH

3 0
3 years ago
Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1
NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x                       x               x

The dissociation constant from the above reactions is given by:

Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}

6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}

6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75    

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

\% = \frac{x}{[HCN]} \times 100

\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100

\% = 3.5 \cdot 10^{-3} \%          

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!      

6 0
2 years ago
Looking at the same nonmetal group on the periodic table, how does the reactivity of an element in period 2 compare to the react
N76 [4]

Answer:

The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.

Explanation:

Nonmetals want to add more electrons to their valence electron shell. Elements in period 2 have their valence shell closer their nucleus, then they are more reactive than period 4 elements.

6 0
3 years ago
A 45g Aluminum spoon (specific heat 0.88J/g degree Celcius) at 24 degrees Celcius placed in 180ml(180g) of coffee at 85 degrees
yarga [219]

Answer:

82 °C

Explanation:

Let the specific heat capacity of the coffee be that of water which is 4.2 J/g °C.

Now, at the final temperature, heat gained by Aluminum spoon ,Q equals heat lost by coffee, Q'.

Q = -Q'

Q = m₁c₁(T₂ - T₁) where m₁ = mass of aluminum spoon = 45 g, c₁ = specific heat of aluminum = 0.88 J/g °C, T₁ = initial temperature of aluminum spoon = 24 °C and T₂ = final temperature of aluminum spoon.

Q' = m₂c₂(T₂ - T₃) where m₂ = mass of coffee = 180 g, c₂ = specific heat of coffee = 4.2 J/g °C, T₃ = initial temperature of coffee = 85 °C and T₂ = final temperature of coffee.

So, Q = -Q'

m₁c₁(T₂ - T₁) = -m₂c₂(T₂ - T₃)

Making T₂ subject of the formula, we have

m₁c₁T₂ - m₁c₁T₁ = -m₂c₂T₂ + m₂c₂T₃

m₁c₁T₂ + m₂c₂T₂ =  m₂c₂T₃ + m₁c₁T₁

(m₁c₁ + m₂c₂)T₂ =  m₂c₂T₃ + m₁c₁T₁

T₂ =  (m₂c₂T₃ + m₁c₁T₁)/(m₁c₁ + m₂c₂)

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

T₂ =  (180 g × 4.2 J/g °C × 85 °C + 45 g × 0.88 J/g °C × 24 °C )/(45 g × 0.88 J/g °C  + 180 g × 4.2 J/g °C)

T₂ =  (64260 J + 950.4 J)/(39.6 J/°C  + 756 J/°C)

T₂ =  65210.4 J/795.6 J/°C

T₂ =  81.96 °C

T₂ ≅  82 °C

8 0
3 years ago
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