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Norma-Jean [14]
2 years ago
15

What is the balanced form of the chemical equation shown below?

Chemistry
1 answer:
Serga [27]2 years ago
4 0

Answer:

A; 1 C_{12}H_{22}O_{11} + 12 O_{2}  - > 11H_2O +12CO_2

Explanation:

The best way to start solving this problem is to start with the molecule with the most atoms. Since there are 12 carbons on the left, you need 12 on the right so 12 would need to be placed in front of carbon dioxide. Also you need 22 hydrogens and in each molecule of water, there are two hydrogen molecules so you need 11 molecules of water. After balancing you find that you need 24 oxygen on the left so you place the coeffecient 12 in front of the oxygen molecule.

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Assume that 1.0 mol of C4H10 is completely burned in excess oxygen to form carbon dioxide and water. How many moles of CO2 would
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You need to first write a chemical equation and balance it
 C₄H₁₀ + O₂ → CO₂ + H₂O
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
1.0 moles               X moles
1.0 mol C₄H₁₀ (\frac{8 mol CO₂}{2 mol C₄H₁₀}) = 4 moles of CO₂
7 0
3 years ago
30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at
Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0
2 years ago
Which state of matter has the most energy?
Softa [21]
It’s gas because the molecules are moving quicker.
8 0
2 years ago
Read 2 more answers
The hydroxide ion concentration of a solution is 0.00001 moles per liter. what is the hydrogen ion concentration, and is the sol
kirza4 [7]
Answer is: the hydrogen ion concentration is 10⁻¹⁰ mol/L, solution is basic.
[OH⁻] =0.00001 mol/L = 10⁻⁴ M.
[OH⁻]·[H⁺<span>] = Kw.
</span>0.00001 mol/L ·[H⁺] = 10⁻¹⁴ mol²/L².
[H⁺] = 10⁻¹⁴ mol²/L²÷ 10⁻⁴ mol/L.
[H⁺] = 10⁻¹⁰ mol/L.
pH = -log[H⁺].
pH = -log(10⁻¹⁰ mol/L).
pH = 10.
If pH is less than seven, solution is acidic; greater than seven, solution is basic; equal seven, solution is neutral.

7 0
3 years ago
What is the value of the rate constant k for this reaction?When entering compound units, indicate multiplication of units explic
vekshin1

The table with the data is in the picture attached.

Answer:

  • k=0.0033M^{-2}.s^{-1}

Explanation:

The reaction equation suggests that the law could have this form:

  • rate=k[A]^a[B]^b[C]^c

Then, the work is to find the values of the exponents that satisfy the initial rate data.

A first glance shows that for the third and fourth trials the initial rates are the same. Since for these two trials only the initial concentration of substance B changed (A and C were kept equal), you conclude that the reaction rate does not depend on B, and ist exponent (lower b) is 0.

Then, so far you can say:

  • rate=k[A]^a[C]^c

When you use trials 1 and 2, you get:

\frac{r_2}{r_1}=\frac{27M/s}{9M/s}=\frac{(0.3M)^a(0.3M)^b(0.9M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=3^c\\\\ 3=3^c\\ \\ 1=c

Now, you can use trials 1 and 3 to determine the other exponent:

\frac{r_3}{r_1}=\frac{36M/s}{9M/s}=\frac{(0.6M)^a(0.3M)^b(0.3M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=2^a\\\\ 4=2^a\\ \\ 2^2=2^a\\ \\ 2=a

Thus, you have the rate law:

  • r=k[A]^2[C]

Now, you just use any trial to obtain k. Using trail 1:

  • k(0.3M)^2(0.3M)=9.10^{-5}M/s

Which yields:

  • k=0.0033M^{-2}.s^{-1}

4 0
3 years ago
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