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AnnyKZ [126]
3 years ago
13

Write the name of any two metalloids ?​

Chemistry
2 answers:
nika2105 [10]3 years ago
7 0

Answer:

Boron and silicon are metalloids.

Hope it helps :)

sertanlavr [38]3 years ago
4 0

Answer:

Germanium and Polonium

Explanation:

Can you please mark me brainliest since I was the first person to answer :p

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Compounds X and Y are both C7H15Cl products formed in the radical chlorination of 2,4-dimethylpentane. Base-promoted E2 eliminat
saul85 [17]

Answer:

X  is 3-chloro-2,4-dimethylpentane

Explanation:

The radical chlorination of 2,4-dimethylpentane may give three products.

Based on the information that

i) both the products gives single alkene on elimination and

ii) both the products also undergoes SN2 reaction but Y reacts faster than X

We may conclude that as SN2 reaction is faster in primary alkyl halides as compared to secondary or tertiary so Y must be primary or secondary and it cannot be tertiary alkyl halide.

The possible products are shown in the figure.

The structure of X is shown in the figure.

5 0
3 years ago
A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

3 0
4 years ago
What are the three ways that an object can interact with a static field?
Kitty [74]

Answer:

friction, conduction, and induction

Explanation:

4 0
3 years ago
true or false: ionic bonding involves the shareing of electrons, oxide ions have a 2+ charge , electrons have a negative charge
Elis [28]
Ionic bonding does not involve the sharing of electrons, that one is false. In ionic bonding, the metal's electrons are given to the non-metal, so that they're both like the nearest noble gas (full electron shells.) They are then drawn together because one has a negative charge (the non-metal) and one has a positive charge (the metal.)

Oxide ions have a 2+ charge: This is false, oxide ions have a 2- charge.

Electrons do have a negative charge, this is true.

Hydrogen molecule: pretty sure this is true. We know this because both atoms are non-metals.

The last one is true: a covalent bond is a shared pair of electrons between two atoms, however be aware that there can be more than one covalent bond between two molecules.


7 0
3 years ago
Read 2 more answers
The decomposition of NH4HS is endothermic: NH4HS(s)⇌NH3(g)+H2S(g) Part A Which change to an equilibrium mixture of this reaction
shepuryov [24]

Explanation:

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

As the given reaction is as follows.

       NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)

(a)  When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.

Thus, formation of H_{2}S will increase.

  • (b)  When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.

Hence, formation of H_{2}S will decrease with decrease in volume.

  • When we increase the mount of NH_{4}HS then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.

Thus, we can conclude that formation of H_{2}S will increase then.

3 0
3 years ago
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