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gladu [14]
3 years ago
14

NEED ANSWERS NOW 100 PTS

Chemistry
2 answers:
Nezavi [6.7K]3 years ago
6 0

Answer:

Iron III fluoride is ionic

Carbon disulfide is a covalent

fiasKO [112]3 years ago
4 0

Answer:

iron i believe is ionic and carbon is covalent

Explanation:

if im wrong correct me

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Deforestation and inappropriate agriculture otherwise known as over farming are factors of and causes of desertification. 
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Which characteristic cannot be inherited?
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Answer:

Knowledge of facts

Explanation:

All of these are inherited traits: Color, Height, and Shape

but Knowledge is a learnt trait

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A rigid plastic container holds 1.00 l methane gas at 660 torr pressure when the temperature is 22.0 degrees Celsius. How much p
igor_vitrenko [27]
 <span>PV/T = P'V'/T' 
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6 0
2 years ago
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A 250-mL aqueous solution contains 1.56 10–5 g of methanol and has a density of 1.03 g/mL. What is the concentration in ppm?
Hoochie [10]
First, calculate for the mass of the aqueous solution by multiplying the given volume (in mL) by the density (in g/mL). In mathematical equation, that is,

        m = ρV

where m is mass, ρ is density, and V is volume. Substituting the known values,

      m = (1.03 g/mL)(250 mL) = 257.5 g

To get the concentration in ppm, divide the given mass of methanol by the mass of the solution. Note that the parts-per million (ppm) is equal to mass of solute in milligram(mg) divided by the mass of solution in kilogram (kg)

      C (in ppm) = (1.56 x 10^-6 g)(1000 mg/1 g)   / (257.5 g)(1 kg/1000 g)

Simplifying,
     C (in ppm) = (1.56 x 10^-3 mg)/ 0.2575 kg

     C (in ppm) = 0.00606 ppm

<em>Answer: 0.00606 ppm</em>
     
7 0
2 years ago
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Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super
dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below:X^{m+}Y^{n-}\rightarrow X_nY_m

Thus, for all the given combinations, we obtain:

- Y⁻

X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3

- Y²⁻

X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3

- Y³⁻

X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY

Best regards!

8 0
2 years ago
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