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3 years ago

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Why is cyanide ion negatively charged?

1 answer:

Bogdan [553]3 years ago

8 0

The cyanide is $CN^{-}$
A carbon atom has 4 valance electrons and nitrogen has 5. Below is a Lewis-dot-structure of cyanide.
:N≡C.
The carbon atom is still one electron short of having a full octet and so it will seize another electron from almost anything, making the cyanide ion negative and whatever it took the electron from it now positive.

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What is the molarity of a stock solution if 60 mL were used to make 150 ml<br>of a .5M solution?

olga2289 [7]

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $$\frac{V1M1}{V2}$

Plugin the values as,

M2 = $$\frac{150 \times 0.5}{60}$

= 1.25 M

So the molarity of the stock solution is 1.25 M.

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Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many

Flauer [41]

Answer : The amount of formaldehyde permissible are, $5.4\times 10^{-6}g$

Explanation : Given,

Density of air = $1.2kg/m^3=1.2g/L$ $(1kg/m^3=1g/L)$

First we have to calculate the mass of air.

$\text{Mass of air}=\text{Density of air}\times \text{Volume of air}$

$\text{Mass of air}=1.2g/L\times 6.0L$

$\text{Mass of air}=7.2g$

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = $\frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}$

Amount of formaldehyde in 7.2 g of formaldehyde = $7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}$

Amount of formaldehyde in 7.2 g of formaldehyde = $5.4\times 10^{-6}g$

Thus, the amount of formaldehyde permissible are, $5.4\times 10^{-6}g$

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3 years ago

When a candle is cooling down is it a physical or chemical change

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Physical

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3 years ago

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Thepotemich [5.8K]

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How many moles of argon gas would be present in a 37.0 liter vessel at 45.00 °C at a pressure of 2.50 atm?

Pani-rosa [81]

Answer:

3.54 mol

Explanation:

Step 1: Given data

Volume (V): 37.0 L

Temperature (T): 45.00 °C

Pressure (P): 2.50 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 45.00°C + 273.15 = 318.15 K

Step 3: Calculate the number of moles (n) of argon gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 2.50 atm × 37.0 L/(0.0821 atm.L/mol.K) × 318.15 K = 3.54 mol

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3 years ago