The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

baherus [9]

Explanation:

Formula for work done is as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now, we will put the given values into the above formula and calculate work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is $3.68 \times 10^{-18} J$ .

7 0

2 years ago

Solid aluminum metal and diatomic chlorine gas react spontaneously to form a solid product. Give the balanced chemical equation

ValentinkaMS [17]

When solid aluminum metal is reacted with diatomic chlorine gas, solid aluminum chloride is formed. This reaction is an example of synthesis or chemical combination in which two elements, aluminum and chlorine combine to form a new compound aluminum chloride.

Word equation: Aluminum (s)+ Chlorine (g)---> Aluminum chloride(s)

Molecular formula of the product formed is $AlCl_{3}$ .

Therefore the balanced chemical equation representing the reaction of solid aluminum with gaseous dichlorine can be represented as,

$2Al(s) + 3Cl_{2}(g)-->2AlCl_{3}(s)$

8 0

2 years ago

Express each aqueous concentration in the unit indicated.

MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L = 9500 millileters (ml)

we consider 25 mg = 25 millileters

<em>ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)</em>

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml = 10000 × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

<em>ppm = (mass of solute / mass of solvent) × 1000000 (1 million)</em>

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

<em>moles = mass / molar mass</em>

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

<em>Molarity = mol of solute / liter of solution</em>

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

<em>moles = mass / molar mass</em>

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

<em>Molarity = moles of solute / litres of solution</em>

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0

2 years ago

a student adds 3.5 moles of solute to enough water to make a 1500mL solution. what is the concentration?

aksik [14]

<h2>Hello!</h2>

The answer is:

$MolarConcentration=\frac{3.5moles}{volume(1.5L)}=2.33molar$

Since there is not information about the solute but only its mass, we need to assume that we are calculating the molar concentration of a solution or molarity. So, need to use the following formula:

$MolarConcentration=\frac{mass(solute)}{volume(solution)}$

Now, we know that the mass of the solute is equal 3.5 moles and the volume is equal to 1500 mL or 1.5L

Then, substituting into the equation, we have:

$MolarConcentration=\frac{3.5moles}{1.5L}=2.33molar$

Have a nice day!

7 0

2 years ago

Read 2 more answers

The atoms of a certain element each contain 19 protons and 1 valence electron. Which statement correctly identifies this element

Leya [2.2K]

Answer

its a potassium elements

Explanation:

K 2,8,8,1 period num 4

group 1A

or 1s1 ,2s2 2p6, 3s1

its a metal reacted rapidly with water to form a colorless basic solution of potassium hydroxide (KOH) and hydrogen gas (H2). The reaction continues even when the solution becomes basic. The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic.

3 0

1 year ago