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Reil [10]
3 years ago
13

Reducing the volume of a contained gas by one third, while holding temperature constant, causes pressure to A. be decreased by t

wo thirds. B. be increased by two thirds. C. be decreased by one third. D. be increased by one third
Chemistry
1 answer:
Nostrana [21]3 years ago
5 0
Reducing the volume of contained gas by one third, while holding temperature constant, causes pressure to D. be increased by one third
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The carbon dioxide gas that was generated during this reaction was collected at 295K and 125 kPa. If 43.2 L of carbon dioxidegas
Vladimir [108]

Answer:

The balanced equation for this reaction is C2H2 + 502 + 4H2O + 3C02. What volume of carbon dioxide is produced when 2.8 L of oxygen are consumed? 25Explanation:

8 0
2 years ago
Name the compound Fe(NO2)2?
Contact [7]
The compound Fe(NO2)2 is Iron (II) Nitrite. 
Hope I could help :) 
5 0
3 years ago
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Fill in the coefficients that will balance the following reaction a0h2so4 + a1koh → a2k2so4 + a3hoh
Anuta_ua [19.1K]
A0 = 1
a1 = 2
a2 = 1
a3 = 2

This can be solved by guessing and checking and making sure that all atoms on the left side are accounted for on the right side. Both sides must have the same amount of atoms or the balancing is not correct.
7 0
3 years ago
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How many grams of water vapor (H2O) are in a 10.2-liter sample at 0.98 atmospheres and 26°C?
diamong [38]
Hey there! 

<span>Use the equation of Clapeyron:
</span>
T in kelvin :

26 + 273.15 => 299.15 K

R = 0.082

V = 10.2 L

P = 0.98 atm

number of moles :

P *V = n * R * T

0.98 * 10.2 =  n * 0.082 * 299.15

9.996 = n * 24.5303

n = 9.996 / 24.5303

n = 0.4074 moles

Therefore:

Molar mass H2O = 18.01 g/mol

1 mole H2O ------------- 18.01 g
0.4074  moles ----------- m

m = 0.4074 * 18.01 / 1

m = 7.339 g of H2O
5 0
3 years ago
How many molecules of co2 in 97.3 grams of co2
s2008m [1.1K]

Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

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

Explanation:

<u>Step 1: Define</u>

97.3 g CO₂

<u>Step 2: Define conversions</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>

97.3 \ 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} ) = 1.33138 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

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

1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂

8 0
3 years ago
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