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Rzqust [24]
3 years ago
12

Water quality tests show the presence of a single heavy metal in the drinking water pulled from a local river. Residents are suf

fering from numerous stomach ailments and are showing signs of long-term kidney damage. The residents of the next town, down river from the first town, seem to be unaffected. Water quality tests in the river by the second town show two types of heavy metals at acceptable legal limits in the water. One of the metals is the same as the one in the first town. What can scientists infer about the possible reason for the illnesses appearing in the first town but not the second?
One metal causes an acute effect, compared to the other.
One metal causes a chronic effect, compared to the other.
The heavy metals have a synergistic relationship when combined.
The heavy metals have an antagonistic relationship when combined.
Chemistry
1 answer:
just olya [345]3 years ago
4 0

Answer:

One metal causes a chronic effect, compared to the other

Explanation:

because the question only said ONE is the same, didn't they? so that would mean that not all the metals are the same for both towns

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Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
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The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

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Where do the protons and electrons come from that are used in the electron transport chain?
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Answer:

4.78 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em></em>

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of MgSO₄ = 50.0 g,

mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.

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<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>

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