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creativ13 [48]
3 years ago
11

What determines whether an object will scatter light or reflect light?

Chemistry
1 answer:
Masteriza [31]3 years ago
4 0
Other factors that affect how much light<span> is absorbed, reflected and </span>scattered concerns<span> the </span>light<span> itself and how it arrived at the surface of the </span>object<span>, such as the wavelength (color) of the </span>light<span>, and. the angle at which it reaches the surface.</span>
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Will NaCl be soluble or insoluble
attashe74 [19]

NaCl is salt to it is obviously SOLUBLE :)

6 0
3 years ago
How is mercury obtained and separated from other nearby substances?​
VLD [36.1K]

Answer:

Most mercury forms in a sulfide ore called cinnabar, but mercury is also frequently found in small amounts in other ores. A common method for separating mercury from cinnabar is to crush the ore and then heat it in a furnace in order to vaporize the mercury. This vapor is then condensed into liquid mercury form.

Explanation:

7 0
2 years ago
Read 2 more answers
Given 0.02 of KF, how many liters of solution are needed to make a 7.2 solution?
Dmitry_Shevchenko [17]

Answer:

1/360

Explanation:

let x = liters

molarity=moles of solute/liters of solution, 7.2=0.02/x or 7.2=(1/50)(1/x), 7.2(50)=(1/x), 360(x)=1, x=1/360

7 0
3 years ago
Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y
olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

\lambda =\frac{0.693}{t_{\frac{1}{2}}}

A=A_o\times e^{-\lambda t}

Where:

\lambda= decay constant

A_o =concentration left after time t

t_{\frac{1}{2}} = Half life of the sample

Half life of Pu-239 = t_{\frac{1}{2}}=24,000 y[

\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]

Let us say amount present of  Pu-239 today = A_o=x

A = ?

A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}

A=0.9715\times x

\frac{A}{A_0}=\frac{A}{x}=0.9715

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0
3 years ago
Describe the overall enthalpy of the chemical reactants compared to the enthalpy of the chemical products in the combustion of p
Veronika [31]

Answer:

A thermochemical equation for the combustion of propane (C3H8)(C3H8) is written as follows:

C3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔH∘rxnC3H8(l)+5O2(g)→3CO2(g)+4H2O(g);ΔHrxn∘ = -2202.0 kJ/mol

The value given for ΔH∘rxnΔHrxn∘ means that:

a. the reaction of one mole of propane absorbs 2202 kJ of energy from the surroundings.

b. the reaction is endothermic.

c. the enthalpy of formation of propane is 2202 kJ/mol.

d. the reaction of one mole of propane releases 2202 kJ of energy to the surroundings.

e. None of these.

3 0
1 year ago
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