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Zinaida [17]
3 years ago
14

How many moles of mercury are equivalent to 3.46 x 1023 atoms?

Chemistry
1 answer:
olga55 [171]3 years ago
4 0

Answer:

3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.

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A mixture contains NaHCO3 together with unreactive components. A 1.54 g sample of the mixture reacts with HA to produce 0.561 g
Marianna [84]

Answer:

69.55 (w/w) %

Explanation:

When NaHCO3 reacts with an acid HA, the reaction that occurs is:

NaHCO3 + HA → H2O + NaA + CO2

<em>Where 1 mole of NaHCO3 produce 1 mole of CO2</em>

<em />

Thus, we need to convert the mass of CO2 to moles using its molar mass (44g/mol). Then, based on the chemical equation, moles of CO2 produced are equal to moles of NaHCO3 in the mixture. With its molar mass -84g/mol- we can find the mass of NaHCO3 and mass percent:

<em>Moles CO2:</em>

0.561g * (1mol / 44g) = 0.01275 moles CO2 = Moles NaHCO3.

<em>Mass NaHCO3:</em>

0.01275 moles * (84g/mol) = 1.071g NaHCO3

<em>Mass percent:</em>

1.071g NaHCO3 / 1.54g sample * 100

<h3>69.55 (w/w) %</h3>

8 0
3 years ago
How do i make a friend?
Diano4ka-milaya [45]
Ask someone to be your friend. If they dont want to be your friend then ask someone else. 

3 0
3 years ago
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I need help with that question please help me I really need it so much I been struggling and stressing a lot please it is due to
s2008m [1.1K]

Answer:

5000*C abd 1.5 million atmospheres

6 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
Please help me with this
Sophie [7]

Answer:

5.0 38 84.0 749.7 528.0 729.0 738.9 739.0

7 0
2 years ago
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