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jek_recluse [69]
3 years ago
8

A dioxin-contaminated water source contains 0.083% dioxin by mass. How much dioxin is present in 3.3 L of this water? Assume a d

ensity of 1.00 g/mL
Chemistry
1 answer:
KatRina [158]3 years ago
3 0

Answer:

2.739 grams of dioxin is present in 3.3 L of this water.

Explanation:

Volume of water , V = 3.3 L = 3.3 × 1000 mL = (1 L = 1000 mL)

Mass of water = m

Density of water  =d = 1.00 g/mL

d=\frac{m}{V}

m=d\times V=1.00 g/mL\times 3.3\times 1000 mL=3300 g

Percentage of dioxon in water = 0.083%

Let the mass of dioxon present in 3300 grams of water be = x

0.083\%=\frac{x}{3300 g}\times 100

x=\frac{0.083}{100}\times 3300 g=2.739 g

2.739 grams of dioxin is present in 3.3 L of this water.

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Explanation:

Normal moles of Na^{+} = volume × normal concentration

                              = 4.7 × 0.139 = 0.6533 mol

Moles of Na^{+} in hyponatremia blood = volume × hyponatremia concentration

                              = 4.7 × 0.116 = 0.5452 mol

Moles of NaCl to be added = moles of extra Na^{+} needed

                            = 0.6533 mol - 0.5452 mol = 0.1081 mol

Mass of NaCl = moles × molar mass of NaCl

                        = 0.1081 mol × 58.443

                        = 6.317g

                        = 6.32 g (approx)

Thus, we can conclude that mass of sodium chloride would need to be added to the blood is 6.32 g.                    

3 0
3 years ago
Two electrons are found in the same atom. One has the quantum number set (2, 1, -1, +½), and the other has the quantum number se
SCORPION-xisa [38]

Answer is: same orbital, but have different spin directions.

The principal quantum number (n) is one of four quantum numbers which are assigned to each electron in an atom to describe that electron's state.  

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l = 0, 1.  

The azimuthal quantum number determines its orbital angular momentum and describes the shape of the orbital.  

2) magnetic quantum number (ml) can be ml = -l...+l.  

ml = -1, 0,+1.  

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3) the spin quantum number (ms), is the spin of the electron.  

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If 8.6 L of O2 reacted with excess H2 at STP, what is volume of the gaseous water collected? @h2 (g) + O2 (g) = 2H2 O (g)
LekaFEV [45]
Note that this is occurring at STP, where 22.4L of any gas is equal to 1mol of that gas.

First, convert the liters of O₂ to moles of O₂ using the conversion factor 22.4LO₂ = 1molO₂.
8.6LO₂ × 1molO₂/22.4LO₂
= 8.6/22.4
≈ 0.3839molO₂

Next, convert moles of O₂ to moles of H₂O. In the balanced equation, the coefficients show that there are 2 moles of H₂O for every mole of O₂. So, use the conversion factor 1molO₂ = 2molH₂O.
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Finally, convert the moles of H₂O to liters of H₂O using the same conversion factor from before, 22.4LH₂O = 1molH₂O.
0.7678molH₂O × 22.4LH₂O/1molH₂O
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≈ 17LH₂O

So, the answer is 17 liters of gaseous water is collected! Note that its rounded to 17 because the measurement given in the problem has 2 sig figs. Hope that helps! :)
3 0
3 years ago
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