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ozzi
3 years ago
10

If an element is lustrous, brittle, and a semi-conductor, how would you classify it?

Chemistry
1 answer:
just olya [345]3 years ago
7 0

Metalloid

Explanation:

If an element is lustrous, brittle and a semi-conductor, it is best classified as a metalloid.

Metalloids shares attributes of metals and non-metals.

  • They are often described as semi-metals as they do not share the full properties that makes a metal a metal.
  • Metalloids are lustrous but not malleable like metals.
  • They do not conduct electricity but they do so on certain conditions.
  • Examples are silicon, germanium, boron, arsenic e.t.c
  • They are usually found in the middle of the periodic table.
  • They are not readily alloyed with metals.

Learn more:

Metalloid brainly.com/question/3023499

#learnwithBrainly

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Explain why landscapers might sift soil<br> that they use to construct gardens.
Studentka2010 [4]

Answer:

When one desires to remove debris from the garden then a soil sifter can be used as a beneficial tool. Based on the needs, it can be of different kinds. A landscaper may sift soil that they use at the time of constructing gardens due to many reasons:  

1. The soil becomes aerated, and thus, turn soft and easy to work upon.  

2. Sifting the soil makes the work of withdrawing undesired substances from the soil easy.  

3. It helps the plants to grow much better as the roots possess the tendency to penetrate more easily through the soil.  

4. The soil becomes healthy due to shifting, thus, helps in producing a beautiful and healthy landscape.  

6 0
3 years ago
Hydrogen produced from a hydrolysis reaction was collected over water and the following data was compiled.
Shkiper50 [21]

Answer:

  • 0.00358 mol

Explanation:

<u>1) Data:</u>

a) V = 93.90 ml

b) T = 28°C

c) P₁ = 744 mmHg

d) P₂ = 28.25 mmHg

d) n = ?

<u>2) Conversion of units</u>

a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter

b) T = 28°C = 28 + 273.15 K = 301.15 K

c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm

d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm

<u>3) Chemical principles and formulae</u>

a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.

b) Ideal gas equation: pV = nRT

<u>4) Solution:</u>

a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm

b) Moles of hygrogen gas:

pV = nRT ⇒ n = pV / (RT) =

n =  (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =

n = 0.00358 mol (which is rounded to 3 significant figures) ← answer

7 0
3 years ago
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Which type of power generation has resulted in raising the levels of the green house effect ?
Eduardwww [97]
A....fossil fuels!
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3 0
3 years ago
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The EPA scientist measures the pH level in one area of a river to be lower than normal and writes that a pollutant must have bee
shutvik [7]
Its an example of inference. 
4 0
3 years ago
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Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo
Romashka-Z-Leto [24]

Answer:

\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

Explanation:

Volume of a cone:

  • \displaystyle V=\frac{1}{3} \pi r^2 h

We have \displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec} and we want to find \displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \ ? when the height is 2 cm.

We can see in our equation for the volume of a cone that we have three variables: V, r, and h.

Since we only have dV/dt and dh/dt, we can rewrite the equation in terms of h only.

We are given that the height of the cone is 1/5 the radius at any given time, 1/5r, so we can write this as r = 5h.

Plug this value for r into the volume formula:

  • \displaystyle V =\frac{1}{3} \pi (5h)^2 h  
  • \displaystyle V =\frac{1}{3} \pi \ 25h^3

Differentiate this equation with respect to time t.

  • \displaystyle \frac{dV}{dt}  =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}
  • \displaystyle \frac{dV}{dt}  =25 \pi h^2 \ \frac{dh}{dt}

Plug known values into the equation and solve for dh/dt.

  • \displaystyle 10 = 25 \pi (2)^2  \ \frac{dh}{dt}
  • \displaystyle 10 = 100 \pi  \ \frac{dh}{dt}  

Divide both sides by 100π to solve for dh/dt.

  • \displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}
  • \displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}

The height of the cone is increasing at a rate of 1/10π cm per second.

7 0
3 years ago
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