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Zarrin [17]
3 years ago
11

The mass of 900. atoms of sodium(Na)

Chemistry
1 answer:
ivanzaharov [21]3 years ago
7 0

Answer:

3.44 × 10⁻²⁰ g Na

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

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

Explanation:

<u>Step 1: Define</u>

900 atoms Na

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

Avogadro's Number

Molar Mass of Na - 22.99 g/mol

<u>Step 3: Convert</u>

<u />900 \ atoms \ Na(\frac{1 \ mol \ Na}{6.022 \cdot 10^{23} \ atoms \ Na} )(\frac{22.99 \ g \ Na}{1 \ mol \ Na} ) = 3.4359 × 10⁻²⁰ g Na

<u>Step 4: Check</u>

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

3.4359 × 10⁻²⁰ g Na ≈ 3.44 × 10⁻²⁰ g Na

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The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
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3 years ago
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Answer:

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3 years ago
When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea
Elan Coil [88]

Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0
3 years ago
The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.
user100 [1]

Answer:

c    43.38 g

Explanation:

The reaction  between MnO2 and HCl can be represented by the following balanced equation:

MnO2 + HCl ---> Cl2 + MnCl2 + H2O

From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.  

For the fact that HCl  gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess,  while the limiting reagent should be MnO2 .  

Thus, the theoretical yield of Cl2 will be  60.25 g.

By definition, the percentage yield is given by

% Yield = (Actual Yield) / (Theoretical Yield),  

This can be simplified to

Actual Yield = % Yield * Theoretical Yield

Plugging in the given values we have

Actual Yield = 72% *  60.25 = 43.38 g

5 0
3 years ago
Can you add more points to question pls
Zarrin [17]

what do you mean? explain what you r trying to dsay


3 0
3 years ago
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