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tresset_1 [31]
3 years ago
14

A sample of gas contains 0.1100 mol of N2(g) and 0.3300 mol of H2(g) and occupies a volume of 20.5 L. The following reaction tak

es place: N2(g) 3 H2(g) 2 NH3(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. g
Chemistry
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

20.5 L

Explanation:

-The stoichiometric reaction of the 2 gases is:

N_2_{(g)}+3H_2_{(g)}->2NH_3_{(G)}

-We apply the combined gas law to determine the resultant volume of the ammonia gas formed:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\\\#But\\\\\frac{P_1}{T_1}=\frac{P_2}{T_2}\\\\\therefore V_1=V_2

Since, the temperature and pressure remain constant, the volume of the ammonia gas formed is equal to the volume of the reactants.

Hence, the volume after the reaction is 20.5 L

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What is a liquid mixture whose parts are evenly blended?.
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Answer:

A homogeneous mixture is a type of mixture in which the individual substances are evenly mixed.

Explanation:

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2 years ago
1. If I have 45 L of He in a balloon at 25 degrees celsius and increase the temperature of the
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Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).

We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.

Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.  

(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).

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3 years ago
Which row in the table shows how many electrons there could be in the outer shell of an atom of V? ​
Amanda [17]

Answer:

Explanation:

The element V forms acidic oxide so it must be non- metal because oxides of non metal forms acidic oxide like sulphur ( s ) or chlorine ( Cl₂ )

sulphur forms SO₂ or SO₃ . Chlorine forms acidic oxide like Cl₂O₇ , Cl₂O₃ etc

These oxides are covalent compounds .

So V may have 6 or 7 electrons in the outermost orbit .

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3 years ago
The flame in a torch used to cut metal is produced by burning acetylene (C2H2) in pure oxygen. Assuming the combustion of 1 mole
Nookie1986 [14]

Answer:

19.8 kg of C₂H₂ is needed

Explanation:

We solve this by a rule of three:

If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene

95.5×10⁴ kJ of heat may be released by the combustion of

(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂

Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g

If we convert the mass from g to kg →  19848 g . 1kg / 1000g = 19.8 kg

7 0
3 years ago
Read 2 more answers
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M
Fed [463]

<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>

What is benzoic acid found in?

  • Some natural sources of benzoic acid include: ​Fruits:​ Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
  • ​Spices:​ Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 =  0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

pH = pKa + log\frac{base}{acid}

4.2 + log\frac{0.05}{0.045} = 4.245

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

pH = pKa + log\frac{base}{acid}

4.245 = 3.75 + log\frac{base}{acid}

log\frac{base}{acid} = 0.5

\frac{base}{acid} = 3.162

Now we must solve the equation above. This will be done using the following values

\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

brainly.com/question/24052816

#SPJ4

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1 year ago
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