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3 years ago

05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises

Chemistry

1 answer:

OLga [1]3 years ago

6 0

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

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S(s)+3F2(g)->SF6(g) how many mol of F2 are required to react completely with 2.30 mol of S?

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Answer: There are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

Explanation:

The given reaction equation is as follows.

$S(s) + 3F_{2}(g) \rightarrow SF_{6}(g)$

Here, 1 mole of S is reaction with 3 moles of $F_{2}$ which means 1 mole of S requires 3 moles of $F_{2}$ .

Therefore, moles of $F_{2}$ required to react completely with 2.30 moles S are calculated as follows.

$1 mol S = 3 mol F_{2}\\2.30 mol S = 3 mol F_{2} \times 2.30 \\= 6.9 mol F_{2}$

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3 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

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Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

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3 years ago

What is the volume of 40.0 grams of argon gas at STP ?

MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up: $\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 24.9235 \ L \ Ar$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

24.9235 L Ar ≈ 24.9 L Ar

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