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Likurg_2 [28]
2 years ago
9

What is the percent composition of calcium fluoride

Chemistry
1 answer:
vladimir1956 [14]2 years ago
8 0

Answer:

78.07

Explanation:

them seperatly is

Calcium Ca 51.333%

Fluorine F 48.667%

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Please help ASAP! I’m lost right now.
SCORPION-xisa [38]
1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1

Chromium is strange because it moves on to the 4s orbital instead of filling the 3d orbital with that last electron. Tricky.

Mark as brainliest if this helped! :)
3 0
3 years ago
Sugar dissolves readily in water because it is a(n) ____ substance.
Mnenie [13.5K]
The answer to this question would be D. Hydrophilic.

The word hydrophilic mean attracted by water. That means the molecule has a force to attract water molecule, thus be able to dissolve in water.  The polarity of the molecule would determine whether a molecule hydrophilic or not.
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3 years ago
What is wrong with the following explanation of an experiment? When it was heated, the sample of carbon decomposed.
Drupady [299]
"Carbon" is an element. It is found in the fourth group of the periodic table, and it is a stable element. This means that it can not be decomposed via heating, because if an element were to break down, it would release its subatomic particles. The explanation was probably one used to describe the thermal decomposition of a compound into smaller compounds.
7 0
3 years ago
What's the Balance equation for:<br>_C4H10O +_ O2---&gt;_CO2+_H20​
steposvetlana [31]

Answer:

C₄H₁₀O + 6O₂ ⇒ 4CO₂ + 5H₂O

Explanation:

Match the amount of reactants and products on both sides of the equation.

3 0
3 years ago
Realiza los cálculos para determinar la cantidad de KOH 90%, que se necesita para preparar 100 ml de solución 1N.
attashe74 [19]

Answer:

6.23 KOH 90% son necesarios

Explanation:

Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.

Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:

<em>Equivalentes KOH:</em>

0.100L * (1eq / L) = 0.100eq = 0.100moles

<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>

0.100moles * (56.1056g/mol) = 5.61 KOH se requieren

<em>KOH 90%:</em>

5.61g KOH * (100g KOH 90% / 90g KOH) =

<h3>6.23 KOH 90% son necesarios</h3>
8 0
2 years ago
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