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Leto [7]
3 years ago
11

When water was added to a 4.00 gram mixture of potassium oxalate hydrate (molar mass 184.24 g/mol) and calcium hydrate shown bel

ow , 1.20 g of calcium oxalate hydrate (146.12 g/mol molar mass) was recovered. If the mole: mole ratio of potassium oxalate to calcium oxalate is 1:1 , what percentage of the 4g-mixture is potassium oxalate
Chemistry
1 answer:
Sliva [168]3 years ago
7 0

Answer:

% (COOK)2H2O = 37.826 %

Explanation:

mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O

∴ mass mix = 4.00 g

∴ mass (CaC2O4)H2O = 1.20 g

∴ Mw (COOK)2H2O = 184.24 g/mol

∴ Mw (CaC2O4)H2O = 146.12 g/mol

∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1

  • % (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100

⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)

⇒ mass (COOK)2H2O = 1.513 g

⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100

⇒ % (COOK)2H2O = 37.826 %

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If a substance undergoes electrolysis and a brown solid forms on one electrode and a gas on the other, from this we can conclude
andrew-mc [135]

Answer:

b. a compound.

Explanation:

Electrolysis is described as a mechanism in which ionic compounds are decomposed into their elements by transmitting a direct electric current via the compound in a liquid state. At the cathode, the cations are reduced and anions at the anode are oxidized. There is an exchange between ions and atoms in the electrolysis process caused by the addition or removal between electrons from the external circuit. As per the question, the original substance is a compound because the electrolysis method is used to obtain pure elements from their respective compound.

7 0
3 years ago
What would be the volume in millilitres of a blood sample of 2.15 microliters (ul)?​
Zanzabum

2.15 x 10⁻³mL

Explanation:

Given parameter:

    Volume of blood sample in uL = 2.15uL

Conversion           uL → mL

   micro- and milli-  are both prefixes of sub-units.

liter is a unit of volume of a substance.

       micro - is 10⁻⁶

       milli- is of the order 10⁻³

The problem is converting from micro to milli:

     if we multiply  10⁻⁶ by 10³ we would have our milli;

  1000uL = 1mL

  2.15uL :   2.15uL x \frac{1mL}{1000uL} = 2.15 x 10⁻³mL

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4 0
3 years ago
How do scientists think that gravity affected the formation of our solar system?
Makovka662 [10]
Maybe because gravity has control of each formation of the solarsydtem thats why its just a guess
5 0
3 years ago
Which statement defines specific heat capacity for a given sample
avanturin [10]

The statement that defines the specific heat capacity for a given sample is the quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

<h3>What is specific heat capacity?</h3>

Specific heat capacity is the of heat to increase the temperature per unit mass.

The formula to calculate the specific heat is Q = mct.

The options are attached here:

  1. The temperature of a given sample is 1 %.
  2. The temperature that a given sample can withstand.
  3. The quantity of heat that is required to raise the sample's temperature by 1 °C1 °C (Kelvin).
  4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

Thus, the correct option is 4. The quantity of heat that is required to raise 1 g of the sample by 1°C (Kelvin) at a constant pressure.

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8 0
1 year ago
A 98.0°C piece of cadmium (c=.850J/g°C) is placed in 150.0g of 37.0°C water. After sitting for a few minutes, both have a temper
Eduardwww [97]
The answer is 19.9 grams cadmium. 
Assuming there was no heat leaked from the system, the heat q lost by cadmium would be equal to the heat gained by the water:
     heat lost by cadmium = heat gained by the water
     -qcadmium = qwater
Since q is equal to mcΔT, we can now calculate for the mass m of the cadmium sample:
     -qcadmium = qwater
     -(mcadmium)(0.850J/g°C)(38.6°C-98.0°C)) = 150.0g(4.18J/g°C)(38.6°C-37.0°C)
     mcadmium = 19.9 grams
3 0
2 years ago
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