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goldenfox [79]
2 years ago
14

What evidence is there from your results that the characteristic color observed for each compound is due to the metal ion in eac

h case, and not the non-metal anion? describe an additional test that could be done to confirm that the color is due to the metal ion?
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
4 0

One part of evidence that the color of the flame created is from the metal ion and not from the chemical is that not any of the flames with dissimilar metals had the similar color (for each metal had its own flame color). Even if most of the metals tested had chloride, the colors of the flames were all dissimilar. The two flames that both had copper (one had copper (II) chloride and the other had copper (II) sulfate) were exactly close in color. The one was green-blue, and the other was a bright green. This displays that they were nearly the same, and the minor difference could be credited to error.

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A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound
Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

          C:H:O  =     1/0.5  :   3/0.5  :   0.5/0.5

          C:H:O  =     2      :     6      :     1

The empirical formula of compound will be C₂H₆O

5 0
2 years ago
One molecule of a compound weighs 2.03x10-22 g. what is the molar mass of that compound?
stealth61 [152]
<span>The molar mass of the compound is 122 g. </span>
6 0
2 years ago
Read 2 more answers
A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calcu
tekilochka [14]

Answer:

x_B=0.0769

m=0.990m

Explanation:

Hello,

In this case, we can compute the mole fraction of benzene by using the following formula:

x_B=\frac{n_B}{n_B+n_C}

Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:

n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol

Thus, we compute the mole fraction:

x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769

Next, for the molality, we define it as:

m=\frac{n_B}{m_C}

Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:

m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg

Or just 0.990 m in molal units (mol/kg).

Best regards.

7 0
2 years ago
Consider the balanced equation below. 4NH3 + 3O2 --&gt; 2N2 + 6H2O What is the mole ratio of NH3 to N2?
AURORKA [14]
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O

From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.

Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
4 0
2 years ago
Read 2 more answers
How many liters of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa)?
OLga [1]

Using ideal gas equation,

P\times V=n\times R\times T

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=100 kPa

T=293 K

R=8.314472 L kPa K⁻¹ mol⁻¹

Number of moles of gas=3.43 mole

Putting all the values in the above equation,

V=\frac{3.43\times 8.314\times 293}{100}

V=83.55 L

So the volume will be 83.55 L.

83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).

4 0
2 years ago
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