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Hatshy [7]
2 years ago
11

What is the pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at

Chemistry
1 answer:
lyudmila [28]2 years ago
3 0

The pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at 145 kPa and 176°C is 1.195 atm.

<h3>What is ideal gas equation?</h3>

Ideal gas equation of any gas will be represented as:
PV = nRT, where

P = pressure

V = volume

n = moles

R = universal gas constant

T = temperature

First we calculate the moles of gas, when the volume of gas 875 ml at

145 kPa and 176°C as:

n = (1.431atm)(0.875L) / (0.082L.atm/K.mol)(449.15K)

n = 1.252 / 36.83 = 0.033 moles

Now we measure the pressure of 0.033 moles of gas of 1.27 L of a gas at 288°C as:

P = (0.033mol)(0.082L.atm/K.mol)(561K) / (1.27L) = 1.195 atm

Hence required pressure of gas is 1.195 atm.

To know more about ideal gas equation, visit the below link:
brainly.com/question/555495

#SPJ1

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How can air pollution affect plants?
GarryVolchara [31]

Answer:

C. air pollution absorbs carbon dioxide

Explanation:

Carbon dioxide's role in the greenhouse effect is a major contributor to air pollution. Radiation and heat emanating from the earth's surface need to be released out into the atmosphere. But because carbon dioxide levels are so high, there is an ozone effect on the ground level.

8 0
3 years ago
If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?
Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

412.072 g Cl₂ ≈ 412 g Cl₂

5 0
3 years ago
How many liters of solvent are needed to make a 6.5M solution with 34 mol of solute?
Dominik [7]

Answer:

5.231 L.

Explanation:

  • Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(Volume of the solution (L))</em>

<em></em>

M = 6.5 M.

no. of moles of solute = 34.0 mol,

Volume of the solution = ??? L.

∴ (6.5 M) = (34.0 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (34.0 mol)/(6.5 M) = 5.231 L.

6 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
saw5 [17]

Answer:

Macroscopic observation

Explanation:

The observation is macroscopic because you could observe the transformation with the naked eye.

3 0
3 years ago
Consider the following reaction between mercury(II) chloride and oxalate ion.
Alina [70]

<u>Answer:</u> The rate law of the reaction is \text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)

Rate law expression for the reaction:

\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b

where,

a = order with respect to HgCl_2

b = order with respect to C_2O_4^{2-}

Expression for rate law for first observation:

3.2\times 10^{-5}=k(0.164)^a(0.15)^b  ....(1)

Expression for rate law for second observation:

2.9\times 10^{-4}=k(0.164)^a(0.45)^b  ....(2)

Expression for rate law for third observation:

1.4\times 10^{-4}=k(0.082)^a(0.45)^b  ....(3)

Expression for rate law for fourth observation:

4.8\times 10^{-5}=k(0.246)^a(0.15)^b  ....(4)  

Dividing 2 from 1, we get:

\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2

Dividing 2 from 3, we get:

\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1

Thus, the rate law becomes:

\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2

3 0
3 years ago
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