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soldi70 [24.7K]
2 years ago
6

Determine the correct set of coefficients to balance the chemical equation. __c6h6(l) + __o2(g)? __co2(g) + __h2o(g)

Chemistry
1 answer:
forsale [732]2 years ago
3 0
To balance this equation, first we should consider balancing C because it only presents in one reactant and one product.  Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2.  Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O.  So the total number of O in the products is 12+3 = 15.  O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O
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Calculate the percent mass of sugar in mountain dew: The total mass of the mountain dew solution is 300g and there are 231 g of
s2008m [1.1K]

Answer:

c. 77 %

Explanation:

Percent mass (% mass) of solute = mass of solute/mass of solution × 100

According to this question, a mountain dew solution weighing 300grams contains 231 g of sugar. This means that:

% mass of sugar = 231g/300g × 100

% mass of sugar = 0.77 × 100

% mass of sugar = 77%.

4 0
2 years ago
A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is
Romashka [77]

Answer:

\large \boxed{\text{B.) 2.8 atm}}

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 1520 Torr; T₁ =   27 °C

p₂ = ?;               T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (  27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\

(c) Convert the pressure to atmospheres

p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}

7 0
2 years ago
Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that
iragen [17]

Answer:

108.43 grams KNO₃

Explanation:

To solve this problem we use the formula:

  • ΔT = Kf * b * i

Where

  • ΔT is the temperature difference (14.5 K)
  • Kf is the cryoscopic constant (1.86 K·m⁻¹)
  • b is the molality of the solution (moles KNO₃ per kg of water)
  • and<em> i</em> is the van't Hoff factor (2 for KNO₃)

We <u>solve for b</u>:

  • 14.5 K = 1.86 K·m⁻¹ * b * 2
  • b = 3.90 m

Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:

  • 275 mL water ≅ 275 g water
  • 275 g /1000 = 0.275 kg
  • moles KNO₃ = molality * kg water = 3.90 * 0.275
  • moles KNO₃ = 1.0725 moles KNO₃

Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:

  • 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
5 0
2 years ago
How many liters of 1.75 M solution could be made using 35 grams of NaCl?
dolphi86 [110]
Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
1.75 =  \frac{35}{58.44*V}
1.75*58.44V = 35
102.27V = 35
V =  \frac{35}{102.27}
\boxed{\boxed{V \approx 0.34\:L}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
How much iodine (I2), in grams, should be added to water to produce 2.5L of solution with a molarity of 0.56M?
denis-greek [22]

Molarity=Moles of solute/Volume of solution in L

So

  • 0.56M=moles/2.5L
  • moles=0.56(2.5)
  • moles of Iodine=1.4mol

Mads of Iodine

  • Moles(Molar mass)
  • 1.4(126.9)
  • 177.66g
7 0
1 year ago
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