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Kazeer [188]
3 years ago
14

If an ideal gas has a pressure of 5.49 atm, a temperature of 88.78 °c, and has a volume of 22.03 l, how many moles of gas are in

the sample?
Chemistry
1 answer:
kirza4 [7]3 years ago
7 0
N = (PV)/RT
(T = 88.78 + 273 = 361.78K)
(R = 22.4/273 = 0.082)
= (5.49 x 22.03)/(0.082 x 361.78) = ?
Put it into the calculator. It's hard to do that on a mobile phone.
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Leya [2.2K]

Answer:

Q=1.04kJ

Explanation:

Hello,

In this case, for latent heat (phase change) we need to consider the enthalpy associated with the involved process, here, melting or fusion; thus, the enthalpy of fusion of copper is 13.2 kJ/mol, therefore, the heat is computed as:

Q=m\Delta H_{fus}

Nevertheless, since the given enthalpy is per mole of copper, we need to use its atomic mass to perform the correct calculation as follows:

Q=5.0g*\frac{1mol}{63.54}* 13.2\frac{kJ}{mol}\\ \\Q=1.04kJ

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Someone answer the limited regent...
OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Using Dimensional Analysis

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table

Explanation:

<u>Step 1: Define</u>

[RxN]   Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given]   10 g Cu

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

[RxN]   1 mol Cu = 1 mol Ag

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Molar Mass of Ag - 197.87 g/mol

<u>Step 3: Stoichiometry</u>

<u />10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} ) = 16.974 g Ag

<u>Step 4: Check</u>

<em>We are given 1 sig fig. Follow sig fig rules and round.</em>

16.974 g Ag ≈ 20 g Ag

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3 years ago
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We know 1 mole of any atom or molecules contains 6.033 \times 10^{23} atom or molecules.

1 mole of HBr i.e 81 gm/mol contains  6.033 \times 10^{23} atom or molecules.

So, mass of 9.03\times 10^{21} molecules is :

m=\dfrac{81\times 9.03\times 10^{21} }{6.033 \times 10^{23}}\\\\m= 1.21\ gm

Therefore, mass of  9.03\times 10^{21} molecules is 1.21 gm .

Hence, this is the required solution.

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