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kifflom [539]
2 years ago
7

True or False? A day is about 12 hours long

Chemistry
1 answer:
7nadin3 [17]2 years ago
7 0

Answer:

false it's 24 hours long.

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Which compound is an Arrhenius base?
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I was hoping that some choices would be given to choose from. As there are no choices given, so i am answering the question based on my knowledge and hope that it comes to your help. Calcium hydroxide is a good example of Arrhenius base. An Arrhenius base is actually a substance that releases a hydroxyl ion in water.
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3 years ago
In a 0.01 M solution of HCl, Litmus will be
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In a 0.01 M solution of HCl, Litmus will be red. Litmus paper will turn into red in acidic conditions. Hydrochloric acid is an acid. Litmus is an indicator for acidity and alkalinity made from inchens.
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Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:_______.
deff fn [24]

Answer:

WCl₂, WCl₄, WCl₅, WCl₆

Explanation:

Molar Mass of Tungsten = 184 g/mol

Mass of Chlorine = 35.5 g/mol

In the first compound;

Percentage of tungsten = 72.17 %

Upon solving;

72.17 % = 184

100 % = Total mass

Total mass of compound = 254.95g

Mass of chlorine = 254.95 - 184 = 70.95 (Dividing by 35.35; This is approximately 2 Chlorine atoms.

The Formular is WCl₂

In the second compound;

Percentage of tungsten = 56.45 %

Upon solving;

56.45 % = 184

100 % = Total mass

Total mass of compound = 325.95 g

Mass of chlorine = 325.95 - 184 = 141.95g (Dividing by 35.35; This is approximately 4 Chlorine atoms.

The Formular is WCl₄

In the third compound;

Percentage of tungsten = 50.91 %

Upon solving;

50.91 % = 184

100 % = Total mass

Total mass of compound = 361.42 g

Mass of chlorine =  361.42 - 184 = 177.42 (Dividing by 35.35; This is approximately 5 Chlorine atoms.

The Formular is WCl₅

In the fourth compound;

Percentage of tungsten = 46.39 %

Upon solving;

46.39 % = 184

100 % = Total mass

Total mass of compound = 396.64 g

Mass of chlorine = 396.64 - 184 = 212.64 (Dividing by 35.35; This is approximately 6 Chlorine atoms.

The Formular is WCl₆

4 0
3 years ago
calculate the volume (in mL) of 0.100 M CaCl2 needed to produce 1.00 g of CaCO3 (s). There is an excess of Na2CO3. Volume of cal
shepuryov [24]

Answer:

100. mL

Explanation:

Step 1: Write the balanced equation for the double displacement reaction

CaCl₂ + Na₂CO₃ ⇒ 2 NaCl + CaCO₃

Step 2: Calculate the moles corresponding to 1.00 g of CaCO₃

The molar mass of CaCO₃ is 100.09 g/mol.

1.00 g × 1 mol/100.09 g = 0.0100 mol

Step 3: Calculate the moles of CaCl₂ required to produce 0.0100 moles of CaCO₃

The molar ratio of CaCl₂ to CaCO₃ is 1:1. The moles of CaCl₂ required are 1/1 × 0.0100 mol = 0.0100 mol.

Step 4: Calculate the volume of 0.100 M CaCl₂ that contains 0.0100 mol

0.0100 mol × 1 L/0.100 mol × 1000 mL/1 L = 100. mL

5 0
2 years ago
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