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ruslelena [56]
1 year ago
12

Why would a biological stain be used? (Give examples of stains and their uses)

Chemistry
1 answer:
Allushta [10]1 year ago
3 0

Answer:

Biological stains are used for the medical and biological industries to aid in detection of structures within tissues. This can include the detection of abnormalities, but is not limited to that. The stains are used to define and examine cell populations within the tissues, to mark cells, or to flag proteins.

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A titration was performed in a lab situation. H2SO4 was titrated with NaOH. The following data was collected: mL of NaOH used =
RoseWind [281]
For the titration we use the equation,
                             M₁V₁ = M₂V₂
where M is molarity and V is volume. Substituting the known values,
                            (0.15 M)(43.2 mL) = (2)(M₂)(20.5 mL)
We multiply the right term by 2 because of the number of H+ in H2SO4. Calculating for M₂ will give us 0.158 M. Thus, the answer is approximately 0.16M. 
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reddi tstudent has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact me
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Explanation:

A point of temperature at which both solid and liquid state of a substance remains in equilibrium without any change in temperature then this temperature is known as melting point.

For example, melting point of water is 0 ^{o}C. So, at this temperature solid state of water and liquid state are present in equilibrium with each other.

Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is 215^{o}C then some of the solid will change into liquid state but the temperature will remains the same.  

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The variables _______ and ________ affect the gravitational force between objects.
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What do radio waves and gamma rays have in common?
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Answer: The sequence from longest wavelength (radio waves) to shortest wavelength (gamma rays) is also a sequence in energy from lowest energy to highest energy. ... The energy carried by a radio wave is low, while the energy carried by a gamma ray is high. Different materials can block different types of light.

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Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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