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garik1379 [7]
3 years ago
6

Which statements describe properties that are unique to metalloids? Check all that apply.

Chemistry
2 answers:
kondaur [170]3 years ago
6 0

Answer:

- metalloids are semiconductive.

- metalloids are amphoteric.

Explanation:

OleMash [197]3 years ago
3 0

The answer is:

- metalloids are semiconductive.

- metalloids are amphoteric.

The explanation:

The metalloids are only 6 elements in the periodic table which are boron, silicon, germanium, arsenic, antimony, and tellurium. And its properties are intermediate between the nonmetal and the metal. and their properties are brittle, shiny ,solid. but the properties that are unique to metalloids.

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Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

7 0
3 years ago
Muszę napisać wzór sumatryczny wysyłam załącznik
ioda

Cześć, nie mówię po polsku, więc zrobiłem to w Tłumaczu Google, więc przepraszam, jeśli coś brzmi śmiesznie.

chlor (VII) i tlen - ten wzór to CI_{2} O_{7}, a tlenek chloru jest bezwodnikiem kwasu nadchlorowego.

węgiel i wodór - wzór na węgiel i wodór to CnH2n + 2), jest to związek organiczny, a niższa klasyfikacja jest taka, że jest to również węglowodór aromatyczny.

Mam nadzieję, że to pomoże, błogosławionego i cudownego dnia! :-)

-Cutiepatutie

7 0
3 years ago
An element has a half-life of 30 years. if 1.0 mg of this element decays over a period of 90 years, how many mg of this element
labwork [276]

Answer:

144.6

Explanation:

6 0
3 years ago
A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?
hodyreva [135]
<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V </span>α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K

By applying the formula,
2.00 L / 313 K = V₂ / 303 K                   
                   V₂ = (2.00 L / 313 K) x 303 K
                   V₂ = 1.94 L

Hence, the volume of the balloon at 30.0 ⁰C  is 1.94 L
7 0
3 years ago
A compound of mercury and oxygen is heated in order to decompose the compound. A 4.08 grams sample of mercury oxide upon heating
arlik [135]

Answer:

HgO (empirical formula)

Explanation:

4.08 - 3.78 = 0.3g (oxygen)

(\frac{4.08}{201})   \:  \:  \:  (\frac{0.3}{16} )

0.02 : 0.02

0.02/0.02 : 0.02/0.02

1 : 1 (ratio)

HgO ( empirical formula)

2HgO ----> 2Hg + O2 ( your equation correct)

7 0
2 years ago
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