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Usimov [2.4K]
2 years ago
14

How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degree

s Celsius and 1.02 atmospheres? CS2(l) + 3O2(g) yields CO2(g) + 2SO2(g) 2.78 liters 5.95 liters 11.9 liters 12.2 liters
Chemistry
1 answer:
alekssr [168]2 years ago
8 0
37.8 grams of CS2 equals to 37.8/76=0.5 mole. So the products have 0.5 mole CO2 which is 11.2 liters at STP. So according to the gas law, the volume at given condition is 12.4 liters. So the answer should be 12.2 liters.
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NEED ASAP THANK YOU!
Bezzdna [24]

Answer:

d. 10

Explanation:

The 24 means the atomic mass and twelve means the number of electrons. You subtract the number of electrons from the atomic mass. 24-12=12

Then, since there's +2, you subtract 2 from 12 which equals 10 which means there are 10 electrons now.

The confusing thing is that + means subtract and - means add here.

Hope this helps.

8 0
2 years ago
Please help Me I really need it
butalik [34]
Can you show the question that goes with those answer pls
7 0
2 years ago
How many moles of glucose does 1.2 x 10^24 molecules represent
sammy [17]
<h3>Answer:</h3>

2.0 mol C₆H₁₂O₆

<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>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 1.2 \cdot 10^{24} \ molecules \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{6.022 \cdot 10^{23} \ molecules \ C_6H_{12}O_6})
  2. Divide:                                                                                                                      \displaystyle 1.99269 \ mol \ C_6H_{12}O_6

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆

3 0
3 years ago
A balloon of helium gas in Alabama has a volume of 22.4 L at 26°C and a 1.03 ATM how big with a balloon get if is transported to
Dimas [21]

Answer:

The final volume of the balloon is = 28.11 L

Explanation:

Initial pressure P_{1} = 1.03 atm = 104.325 K pa

Initial temperature T_{1} = 26 °c = 299 K

Initial volume V_{1} = 22.4 L

Final temperature T_{2} = 22 °c = 295 K  

Final pressure P_{2} = 0.81 atm = 82 K pa

We know that

\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }

Put all the values in above formula we get

\frac{(104.325)(22.4)}{299} = \frac{(82)(V_{2} )}{295}

V_{2} = 28.11 L

This is the final volume of the balloon.

8 0
3 years ago
Calculate the mass in grams for each of the following liquids.
WITCHER [35]
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
3 0
3 years ago
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