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3 years ago

6

A substance contains 36.1 percent calcium and 63.9 percent chlorine by weight. What is the empirical formula of the given compou

nd?
A.
CaCl

B.
Ca0.9Cl1.8

C.
CaCl2

D.
Ca2Cl4

2 answers:

Margaret [11]3 years ago

5 0

Answer:

C

Explanation:

Alika [10]3 years ago

4 0

Answer:C-CaCl2

Explanation:The answer is C because the chemical formula of calcium chloride is CaCl2.

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how do the properties of sodium chloride compare with the properties of its component elements, sodium

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Sodium chloride is a nonreactive solid at room temperature, and is commonly known as table salt. The two elements that make up sodium chloride are sodium and chlorine. Sodium is a very reactive metal that tastes bad. Pure sodium is explosive when it comes in contact with water. Hope this helps

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4 years ago

What is the empirical formula for a compound which contains 67.1 zinc and the rest is oxygen

wolverine [178]

Answer:

The empirical formula is ZnO2

Explanation:

What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

Step 1: Data given

Suppose the compound has a mass of 100.0 grams

A compound contains:

67.1 % Zinc = 67.1 grams

100 - 67.1 = 32.9 % oxygen = 32.9 grams

Molar mass of Zinc = 65.38 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles of Zinc

Suppose the compound is 100 grams

Moles Zn = 67. 10 grams / 65.38 g/mol

Moles Zn = 1.026 moles

Step 3: Calculate moles of O

Moles O = 32.90 grams / 16.00 g/mol

Moles O = 2.056 moles

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

Zn: 1.026/1.026 = 1

O: 2.056/1.026 = 2

The empirical formula is ZnO2

To control this we can calculate the % Zinc for 1 mol

65.38 / (65.38+2*16) = 0.67.1 = 67.2 %

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3 years ago

What does the atomic number of each atom represent?

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How many grams of oxygen are in 5.3 moles of lactose? (Using percent composition)

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3 years ago

The constant volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabaticall

miv72 [106K]

Answer:

The value is $C_p = 42. 8 J/K\cdot mol$

Explanation:

From the question we are told that

$\gamma = \frac{C_p }{C_v}$

The initial volume of the fluorocarbon gas is $V_1 = V$

The final volume of the fluorocarbon gas is $V_2 = 2V$

The initial temperature of the fluorocarbon gas is $T_1 = 298.15 K$

The final temperature of the fluorocarbon gas is $T_2 = 248.44 K$

The initial pressure is $P_1 = 202.94\ kPa$

The final pressure is $P_2 = 81.840\ kPa$

Generally the equation for adiabatically reversible expansion is mathematically represented as

$T_2 = T_1 * [ \frac{V_1}{V_2} ]^{\frac{R}{C_v} }$

Here R is the ideal gas constant with the value

$R = 8.314\ J/K \cdot mol$

So

$248.44 = 298.15 * [ \frac{V}{2V} ]^{\frac{8.314}{C_v} }$

=> $C_v = 31.54 J/K\cdot mol$

Generally adiabatic reversible expansion can also be mathematically expressed as

$P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$

=> $[ 81.840 *10^3] [2V]^{\gamma} = [202.94 *10^3] V^{\gamma}$

=> $2^{\gamma} = 2.56$

=> $\gamma = 1.356$

So

$\gamma = \frac{C_p}{C_v} \equiv 1.356 = \frac{C_p}{31.54}$

=> $C_p = 42. 8 J/K\cdot mol$

3 0

3 years ago