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Sveta_85 [38]
3 years ago
13

A pesticide is used to kill crops in a field. A scientist wants to learn how the pesticide is affecting the plants and animals n

ear the field. Which procedure would be most useful? measuring the radioactivity of a sample of untreated pesticide introducing a radioactive tracer to common compounds in the soil subjecting the pesticide to neutron activation analysis introducing a radioactive tracer into a sample of pesticide
Chemistry
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

introducing a radioactive tracer into a sample of pesticide

Explanation:

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Answer:  mmmmmm asking for mrs.howard work I see lol good luck grace

Explanation:

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3 years ago
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Two substances with empirical formula HNO are hyponitrous acid (M = 62.04 g/mol) and nitroxyl (M = 31.02 g/mol).(a) What is the
viva [34]

The molecular formula for hyponitrous acid is H2N2O2. and for nitroxyl is HNO.

The chemical compound HNO is also known as nitroxyl (common name) or Azanon (IUPAC name). In the gas phase, it is widely recognized. In the solution phase, the short-lived intermediate nitroxyl can develop. Nitric oxide (NO) is reduced to form the conjugate base, NO, which is isoelectronic with dioxygen.

By oxidizing hydroxylamine with CuO , HgO, and Ag 2 and by oxidizing hydroxylamine with N2O3 in methyl-alcoholic solution, we can create hyponitrous acid.

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4 0
2 years ago
5. Which of the following statements is not true?
vladimir1956 [14]

i believe the answer is...

Atoms of isotopes of an element have different numbers of protons.

4 0
3 years ago
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Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/
ankoles [38]

The balanced chemical equation for the combustion of butane is:

2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)

ΔH_{reaction}^{0} = Σn_{products}ΔH_{f}^{0}_{(products)}-Σn_{reactants}ΔH_{f}^{0}_{(reactants)}

                         = [{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

                      = -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

6 0
3 years ago
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