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Alex_Xolod [135]
3 years ago
5

Which of the following statements correctly describe metalloids?

Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
3 0
Metalloids are elements that have both metal and nonmetal properties
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Astrology, a belief in a relationship between constellations of stars and life events,
yanalaym [24]

Answer:

Pseudoscience

Explanation:

"Astrology has not demonstrated its effectiveness in controlled studies and has no scientific validity, thus regarded as pseudoscience."

3 0
3 years ago
As two chemicals react, the chemical bonds between the atoms in each react and break. Then, new bonds form. We call this process
EastWind [94]

Answer:

False is an answer.

8 0
3 years ago
If 5 grams of co and 5 grams of o2 are combined according to the reaction 2co o2 --> 2co2, which is the limiting reagent?
IrinaK [193]

When 5 grams of CO and 5grams of O₂ is combined according to the reaction 2CO + O₂ ----------> 2CO₂ then Limiting reagent is CO.

Limiting reagent is the reactant that get used up in the reaction first.

According to the given reaction:

2CO + O₂ ----------> 2CO₂

5g       5g

∴ Molar mass of CO = Molar mass of C + Molar mass of O

⇒ Molar mass of CO = 14 + 16

⇒ Molar mass of CO = 28g

∴ Molar mass of O₂ = 16(2) = 32g

∴ Molar mass of CO₂ = Molar mass of C + 2(Molar mass of O)

⇒ Molar mass of CO₂ = 14 + 2(16)

⇒Molar mass of CO₂ =44g

Let's find out the moles of CO and O₂

∴ Moles = Given mass / Molar mass

⇒ moles of CO = 5/28 = 0.17

⇒ moles of O₂ = 5/32 = 0.15

For finding out the Limiting Reagent, we will divide the number of moles with the stiochiometry of the given reaction.

⇒ For CO = Moles/ stiochiometry = 0.17/2 = 0.085

⇒ For O₂ = Moles/ stiochiometry = 0.15/1 = 0.15

Since, the ratio of number of moles with the stiochiometry is less for CO hence it is the Limiting reagent, i.e. it will get used up in the reaction first.

Hence, the Limiting reagent  for the reaction is CO.

Learn more about Limiting reagent here, brainly.com/question/11848702

#SPJ4

8 0
1 year ago
In the chemical equation Zn + 2HCl ZnCl2 + H2, the reactants are
Rudik [331]

Answer : The correct option is A.

Explanation :

The given chemical reaction is,

Zn+2HCl\rightarrow ZnCl_2+H_2

In the reaction, Reactants are present on the left side of the arrow and products are present on the right side of the arrow.

Reactants are those which takes part in and undergoes change during the reaction.

Products are those which are formed during the reaction.

In the given reaction, zinc react with hydrochloric acid to form zinc chloride and hydrogen gas.

Therefore, the reactants are zinc and hydrochloric acid.


4 0
3 years ago
Read 2 more answers
Aluminum nitrite and ammonium chloride react to form aluminum chloride, nitrogen, and water. How many grams of each substance ar
Nady [450]

Explanation:

Given Data:

The mass of aluminium nitrite is 72.5 g

The mass of ammonium chloride is 58.6 g

The balanced chemical equation for the reaction is given as follows.

Al(NO2)3 + 3NH4Cl → AlCl3 + 3N2 + 6H2O

The number of moles can be determined by the formula given as follows.

Number of moles = Mass / Molar mass

The molar mass of aluminum nitrate and ammonium chloride is 164.998 g/mol and 53.49 g/mol respectively.

inserting the respective values in the formula given above.

Moles of Al(NO2)3 = 72.5 g / 164.998 g/mol = 0.439 mol

Moles of NH4Cl = 58.6 g / 53.49 g/mol = 1.096 mol

From the balanced equation,

3 moles of ammonium chloride requires 1 mole of aluminum nitrate.

So, 1 mole of ammonium chloride requires 1 / 3 mole of aluminum nitrate.

Thus, 1.096 mole of ammonium chloride will require (1 / 3) × 1.096 = 0.3653 mole of aluminum nitrite.

Here, the amount of aluminum nitrate is more than the required amount so ammonium chloride is the limiting reagent.

From the balanced chemical equation, 3 mole of ammonium chloride gives 1 mole of aluminum chloride.

So, 1.096 mole of ammonium chloride will give;

3 = 1

1.096  = x

x = (1.096 * 1 ) / 3 = 0.3653 mole of aluminum chloride.

Therefore, the number of moles of aluminum chloride is 0.3653 mol.

Since the molar mass of aluminum chloride is 133.34 g/mol

Substitute the respective values in the formula given above.

0.3653 mol = Mass / 133.34 g/mol

Mass = 0.3653 mol × 133.34 g/mol = 48.71 g

Therefore , the mass of aluminum chloride produces is 48.71 g.

The Ammonium chloride is completely used up in the reaction.

The amount of alumium nitrite used is =  Number of moles * Molar mass =  0.3653 * 164.998 = 60.27

Mass of alminium nitrite left = 72.5 - 60.27 = 12.23g

8 0
3 years ago
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