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NikAS [45]
3 years ago
10

22.4 L of hydrogen is taken in a cylinder at zero degree Celsius and at one atm. This gas is transferred to a cylinder of 11.2 L

capacity without change in temperature. The present pressure and mass of the gas is ....... & ......
Chemistry
2 answers:
geniusboy [140]3 years ago
7 0

Answer:

  2 Atm; 2.016 g

Explanation:

Changing the volume without changing the temperature or mass only changes the pressure. Volume and pressure are inversely proportional so halving the volume will double the pressure.

P = 1 Atm, T = 0 °C are "standard" temperature and pressure (STP). The volume of 1 mole of gas is 22.4 L under these conditions. That means the amount of hydrogen gas in the cylinder is 1 mole, so has a mass of 2.016 g.

After the volume reduction, the pressure is 2 Atm, and the mass remains 2.016 g.

GREYUIT [131]3 years ago
7 0

Answer:

<h2>2 ATM ; 2.016 g</h2>

  • <em>After</em><em> </em><em>the</em><em> </em><em>volume</em><em> </em><em>of</em><em> </em><em>reduction</em>

Explanation:

hope it helps

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I believe the problem is just simply asking for us to convert the value from one unit to the other. This case from m^3 to km^3. From the SI units, we know 1 km is equal to 1000 m. We do as follows:

118 m^3 ( 1 km / 1000 m )^3 = 1.18 x 10^-7 km^3
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2_2_2

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Does sodium have a positive or negative charge after ionization?
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3 years ago
1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this
LenKa [72]
<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0
2 years ago
Ammonium chloride is used as an expectorant in cough medicine. It has a density of 1.53 g/cm^3. What is the mass of 26.0L of thi
Katena32 [7]

Answer:

                     39.78 Kg

Explanation:

Data Given;

                  Density  =  1.53 g/cm³

                  Mass  =  <u>??</u>

                  Volume  =  26.0 L  =  26000 cm³

Formula Used;

                       The measure of mass per unit volume is called as density. It is a physical property of a substance and tells how tightly the particles are packed. Mathematically;

                  Density  =  Mass ÷ Volume

Solving for Mass,

                  Mass  =  Volume × Density

Putting values,

                  Mass  =  26000 cm³ × 1.53 g.mL⁻¹

                  Mass  =  39780 g  or   39.78 Kg

8 0
3 years ago
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