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NemiM [27]
3 years ago
15

You dissolve the sample of fd&c red 40 (molar mass 496.4200 g/mol) in enough water to make 250.0 ml of solution. we call thi

s the stock solution. what is the concentration of the stock solution in terms of molarity?
Chemistry
1 answer:
icang [17]3 years ago
8 0
Molar mass 496.4200 g/mol

Number of moles:

40 g x 1 mol / 496.4200 => 0.08057 moles

Volume in liters:

250.0 mL / 1000 => 0.25 L

Therefore:

M = moles / V

M = 0.08057 / 0.25

 = 0.32228 M

hope this helps!
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A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration?
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Answer is 2m/s sq because V-u /t us acceleration 40-10/15=2
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timama [110]

The good ozone protects us from the UV/ harmful radiations whereas bad ozone is an air pollutant.

Explanation:

  • There are two types of ozone layer found in the earth's atmosphere extending from troposphere to stratosphere. They are good ozone and bad ozone.
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7 0
3 years ago
How many moles of CaO will be produced from 23.9 g of Ca?<br>2Ca(s) + O2(g) – 20:0(3)<br>mol​
inessss [21]

Answer:

0.6 moles of CaO will produced.

Explanation:

Given data:

Mass of calcium = 23.9 g

Moles of CaO produced = ?

Solution:

Chemical equation:

2Ca + O₂ → 2CaO

Number of moles of calcium:

Number of moles = mass/ molar mass

Number of moles = 23.9 g / 40 g/mol

Number of moles = 0.6 mol

Now we will compare the moles of calcium and CaO.

            Ca         :          CaO

              2          :          2

            0.6         :        0.6

 0.6 moles of CaO will produced.

4 0
3 years ago
A sample of CO2 weighing 86.34g contains how many molecules?
irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

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>

86.34 g CO₂

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

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

<u />86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.18141 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0
3 years ago
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