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Vera_Pavlovna [14]
2 years ago
8

What are the carbon (C) and oxygen (O) in carbon dioxide (CO2)?

Chemistry
1 answer:
Fynjy0 [20]2 years ago
6 0
A. Atoms. Because I learned this in 6th grade.
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Why does pepsinogen become activated at ph 2?
Katena32 [7]
////// it is pepsinogen //////
5 0
2 years ago
Vitamin C contains only carbon, hydrogen, and oxygen. When a 1.00 g was combusted, 1.4991 g of CO2 and 0.4092 g of H2O were obta
Alina [70]

The empirical formula for this vitamin : C₃H₄O₃

<h3>Further explanation   </h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

  • Determine the mass ratio of the constituent elements of the compound.  
  • Determine the mole ratio by dividing the percentage by the atomic mass

Mass of C in CO₂ :(MW C = 12 g/mol, CO₂=44 g/mol)

\tt \dfrac{12}{44}\times 1.4991=0.409~g

Mass of H in H₂O :(MW H = 1 g/mol, H₂O = 18 g/mol)

\tt \dfrac{2.1}{18}\times 0.4092=0.0455~g

Mass O = Mass sample - (mass C + mass H) :

\tt 1-(0.409+0.0455)=0.5455~g

mol ratio C : H : O =

\tt \dfrac{0.409}{12}\div \dfrac{0.0455}{1}\div \dfrac{0.5455}{16}\\\\0.0341\div 0.0455\div 0.0341\rightarrow 1\div 1.33\div 1=3\div 4\div 3

5 0
3 years ago
An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?
ra1l [238]
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant 
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT 
where 
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation 

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL 
therefore answer is 
c. 448 mL 
5 0
2 years ago
A cylinder is filled with 2.00 moles of nitrogen, 3.00 moles of argon, and 5.00 moles of helium.
Aleks [24]

PRACTICE MULTIPLE CHOICE QUESTIONS FROM UNIT 10

1.) A cylinder is filled with 2.00 moles of nitrogen, 3.00 moles of argon, and 5.00 moles of helium. If

the gas mixture is at STP, what is the partial pressure of the argon?

(A) 152 torr (B) 228 torr (C) 380. torr (D) 760. torr

2.) Compared to the average kinetic energy of 1 mole of water at 0 oC, the average kinetic energy

of 1 mole of water at 298 K is

(A) the same, and the number of molecules is the same

(B) the same, but the number of molecules is greater

(C) greater, and the number of molecules is greater

(D) greater, but the number of molecules is the same

3.) If the pressure on a given mass of gas in a closed system is increased and the temperature

remains constant, the volume of the gas will

(A) decrease (B) increase (C) remain the same

4.) Which gas has approximately the same density as C2H6 at STP?

(A) NO (B) NH3 (C) H2S (D) SO2

5.) At a temperature of 273 K, a 400. milliliter gas sample has a pressure of 760. millimeters of

mercury. If the pressure is changed to 380. millimeters of mercury, at which temperature will

this gas sample have a volume of 551 milliliters?

(A) 100 K (B) 188 K (C) 273 K (D) 546 K

Get the answers to these questions

Back to the Unit 10 Old Tests Page

Back to the Unit 10 Page

Back to the Main Page

5 0
2 years ago
Read 2 more answers
Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to
valina [46]

5.451 X 10³ kg of sodium carbonate must be added to neutralize 5.04×103 kg of sulfuric acid solution.

<u>Explanation</u>:

  • Sodium carbonate is used to neutralized sulfuric acid, H₂SO₄. Sodium carbonate is the salt of a strong base (NaOH) and weak acid (H₂CO₃). The balanced chemical reaction for neutralization is as follows:

                        Na₂CO₃ + H₂SO₄ ----> Na₂SO₄ + H₂CO₃

  • From a balanced chemical equation, it is clear that one mole of Na₂CO₃ is required to neutralize one mole of  H₂SO₄.
  • Molar mass of Na₂CO₃= 106 g/mol = 0.106 kg/mol and                                Molar mass of H₂SO₄= 98 g/mol = 0.098 kg/mol.
  • To neutralize 0.098 kg of H₂SO₄ amount of Na₂CO₃ required is 0.106 kg, so, To neutralize 5.04×10³ kg of H₂SO₄, Na₂CO₃ required is =  5.451 X 10³ kg.

 

5 0
3 years ago
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