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OLEGan [10]
3 years ago
13

Which of these is a mixture of particles that are large enough to scatter or block light

Chemistry
1 answer:
Daniel [21]3 years ago
5 0
The answer is C. because <span>particles settle out over time ,can block light and scatter light .</span>
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Someone help me with this asap for 5p points and BRAINLIEST THANK YOU
GaryK [48]

Answer:

High trainability allows the owner to train the dog to handle certain things--some dog can be trained as police dogs and can be taught various jobs, so high trainability is very important

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Loud bark- A loud bark from a dog can help alert other people when there is a intruder, it makes alot of noise so they can attract attention easily.

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7 0
2 years ago
Anota una alternativa o solución que se pueda realizar para evitar la escasez de agua:
Olin [163]

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3 0
3 years ago
What is the mass of the block of iron illustrated below?
Mandarinka [93]
VOLUME= 5cm*10cm*2cm =100cm^3
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7 0
3 years ago
Read 2 more answers
How many moles of hydrogen ions are formed in the ionization of 0.250 moles of H2SO4?
True [87]

Answer:

The ionization of 0.250 moles of H₂SO₄ will produce 0.5 moles of H⁺ (hydrogen ion)

Explanation:

From the ionization of H₂SO₄, we have

H₂SO₄ → 2H⁺ + SO₄²⁻

Hence, at 100% yield, one mole of H₂SO₄ produces two moles of H⁺ (hydrogen ion) and one mole of SO₄²⁻ (sulphate ion), therefore, 0.250 moles of H₂SO₄ will produce 2×0.250 moles of H⁺ (hydrogen ion) or 0.5 moles of H⁺ (hydrogen ion) and 0.25 moles of SO₄²⁻ (sulphate ion).

That is; 0.250·H₂SO₄ → 0.5·H⁺ + 0.250·SO₄²⁻.

4 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
3 years ago
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