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gtnhenbr [62]
3 years ago
9

A 5 grams sample of liquid water evaporates into gaseous water in a closed container, expanding in the process. What is the mass

after evaporation?
Chemistry
1 answer:
padilas [110]3 years ago
5 0

We have get the mass of gaseous water after evaporation in a closed container.

The mass of water vapor after evaporation is 5 grams.

In closed container, there is no exchange in mass from system to surrounding, only heat may exchange. The number of moles of water vapour remains unchanged as 5 gram water is heated in closed container.

Due to heating, liquid water gets evaporated and intermolecular distance between water molecules increases in gaseous state than liquid state and intermolecuar force of attraction decreases.

Randomness of  molecules increases in gaseous state than liquid state.  


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Name two methods scientists use to obtain empirical evidence
adell [148]
Through hypothesis and experiments
4 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
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
CALCULATOR
suter [353]

Answer: The possible molecular formula will be CO_2

Explanation:

Mass of C= 27.3 g

Mass of O = 72.7 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{27.3g}{12g/mole}=2.275moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{72.7g}{16g/mole}=4.544moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{2.275}{2.275}=1

For O =\frac{4.544}{2.275}=2

The ratio of C : O = 1: 2

Hence the empirical formula is CO_2

The possible molecular formula will be=n\times CO_2

5 0
3 years ago
Well................................
Law Incorporation [45]

Answer:

w t f

w t f

confused to the max

8 0
3 years ago
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