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xxTIMURxx [149]
1 year ago
13

Gamma rays are often used to kill microorganisms in food, in an attempt to make the food safer. Some people contend that this ir

radiated
food is actually less safe to eat. Which, if true, would most effectively dispute their idea? (1 point)
a
Irradiated food only emits alpha particles, which are harmless.
Ob
Irradiated food is shown to not be radioactive.
C
Irradiated food only contributes to background radiation.
Irradiated food has molecules that undergo transmutation,
Od
Chemistry
1 answer:
nikdorinn [45]1 year ago
4 0

Answer:

b . Irradiated food is shown to not be radioactive.

Explanation:

If it can be proven that irradiated food is not radioactive, then it will effective dispute the idea that irradiated food are less safe to eat.

  • An irradiated food is one in which ionizing radiations have been employed to improve food quality.
  • Thus, bacteria and other food spoilers can be exterminated from the food.
  • Most irradiated food do not contain radiation and are fit for consumption.

If it can be proven, that this is true, then it will challenge the idea that irradiated foods are not safe.

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A student took notes in class. - Uses high frequency sound waves - Creates images from echoes - Has many applications - Medical
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The correct option should be ultrasound technology (option B) because it is related to sonographers or ultrasound technicians. they are most likely with while pregnancy but they have plenty of uses, such as evaluating and diagnosis, many medical treatment for elderly patients. 
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A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

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= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

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6 0
2 years ago
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Please help!!
Sati [7]

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

<span>Assuming that there is the same amount  of solution X for each reactant</span>

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

<span>The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.</span>

3 0
2 years ago
A 20.0-mL sample of lake water was acidified with nitric acid and treated with excess KSCN to form a red complex (KSCN itself is
Novosadov [1.4K]

Answer:

8.09x10⁻⁵M of Fe³⁺

Explanation:

Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.

In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-

The concentration of Fe³⁺ in the reference sample is:

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<em>Because reference sample was diluted from 5.0mL to 50.0mL.</em>

<em>That means a solution of  4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512</em>

Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:

0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = <em>3.23x10⁻⁵M.  </em>

As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:

3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = <em>8.09x10⁻⁵M of Fe³⁺</em>

4 0
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