An atom has a diameter of approximately 0.0010 um. How many meters is this?

A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What is the concentration of SO2 in ppm? in ppb?

Mnenie [13.5K]

Answer:

= 230 ppb

Explanation:

Considering that;

1ppm = 1mg/L

Then;

0.00023g = 0.23mg

Therefore;

0.00023 g/L = 0.23 mg/L

0.23 mg/L = 0.23 ppm

1 ppm = 100 ppb

Therefore;

0.23 ppm = 0.23 ×1000

= 230 ppb

5 0

2 years ago

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Prisms separate ________ light, such as that from the Sun, by wavelength. Which best completes the statement? ultraviolet white

tatuchka [14]

Answer:

The answer is B white

Explanation:

hope it helps :)

4 0

2 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains $6.022\times 10^{23}$ atoms

So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

$\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%$

and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0

2 years ago

The potential energy diagram for a reaction starts at 180 kJ and ends at 300 kJ. What type of reaction does the diagram best rep

aleksklad [387]

Answer: Endothermic reaction

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and $\Delta H$ for the reaction comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and $\Delta H$ for the reaction comes out to be positive.

As the energy of reactants is 180 kJ and that of products is 300 kJ, the energy of products is greater than that of reactants, which means the energy has been absorbed and reaction is endothermic.

8 0

2 years ago

50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in th

statuscvo [17]

Answer:

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.

3 0

2 years ago

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