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Nuetrik [128]
3 years ago
7

When 8.21 l of c3h8 (g) burn in oxygen, how many liters of oxygen are consumed? all gas volumes are measured at the same tempera

ture and pressure but not at stp?
Chemistry
1 answer:
just olya [345]3 years ago
5 0
<span>8.21 L of C3H8(g) Lets take c as the molar volume at that temperature.
    c L <><> 5c L
  C3H8 (g) + 5O2 (g) --> 3CO2 + 4H2O + Q
  8.21 L <><> x L
    x = (8.21 * 5c)/c = 8.21 * 5 = 41.05 L O2 consumed for a 100% yield.</span>
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At T = 250 °C the reaction PCl5(g) PCl3(g) + Cl2(g) has an equilibrium constant in terms of pressures Kp = 2.15. (a) Suppose the
Ganezh [65]

Answer:

To the right

Explanation:

Step 1: Given data

  • Partial pressure of PCl₅ (pPCl₅) = 0.548 atm
  • Partial pressure of PCl₃ (pCl₃) = 0.780 atm
  • Partial pressure of Cl₂ (pCl₂) = 0.780 atm

Step 2: Write the balanced equation

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

Step 3: Calculate the pressure reaction quotient

Q_p = \frac{pPCl_3 \times pCl_2 }{pPCl_5} = \frac{0.780 \times 0.780 }{0.548} =1.11

Step 4: Determine whether the reaction proceeds to the right or to the left as equilibrium is approached

Since <em>Qp < Kp</em>, the reaction will proceed to the right to attain the equilibrium.

7 0
3 years ago
Which group will have an electron configuration that ends in s2?
Step2247 [10]
<h2>Answer:   c . Alkaline earth metals (Group 2)</h2>

<h3>Explanation:</h3>

Group 2 metals have 2 electrons in their outer shell. These two electrons are usually found in the s orbital, hence the s².

8 0
2 years ago
Enter your answer in the provided box.
nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, V_1=43.1\ L

Initial temperature, T_1=24^{\circ} C=24+273=297\ K

Final temperature, T_2=78^{\circ} C=78+273=351\ K

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L

So, the final volume of the gas is 50.93 L.

4 0
3 years ago
10) How many grams are there in 1.00 x 10^24 molecules of BC13?
lana66690 [7]

194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

Explanation:

In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.

It is known that 1 moles of any element has 6.022×10²³ molecules.

Then 1 molecule will have \frac{1}{6.022*10^{23} } moles.

So 1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles

Thus, 1.66 moles are included in BCl₃.

Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.

As it is known as 1 mole contains molecular mass of the compound.

As the molecular mass of BCl₃ will be

Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)

Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.

Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.

grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles

grams of BCl_{3}=117.17*1.66=194.5 g

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.

3 0
3 years ago
Initially a beaker contains 225.0 ml of a 0.350 M MgSO4 solution. Then 175.0 ml of water are added to the beaker. Find the conce
Salsk061 [2.6K]
Final volume is 400 mL

<span>The moles in MgSO4 is 0.00788 </span><span>mL
</span>
The new concentration is 0.197
8 0
3 years ago
Read 2 more answers
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