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yuradex [85]
3 years ago
14

Which of these Group 14 elements has the most metallic properties? 1. C 2. Ge 3. Si 4. Sn

Chemistry
2 answers:
Furkat [3]3 years ago
7 0

Answer: The correct answer is option(4).

Explanation:

Metal have property of readily loosing an electron.

When we move down the group , atomic size of the atom increases by which valence electrons move far away from the nucleus. And ability to loose an electron(s) from the valence shell also increases. Hence, metallic character increases on moving down the group.

From the given options of elements of group 14,tin will be last element in the group with the highest metallic character than the carbon , germanium and silicon.

Hence, the correct answer is tin(Sn).

seraphim [82]3 years ago
5 0
I would say that the answer is Sn.
C-is a non-metal
Ge-is a metalliod (consists both non-metal and metal)
Si -is a metalloid
Sn- is a pure metal 
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How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC
sp2606 [1]

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L

Best regards!

4 0
3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
A diploid somatic ("body") cell has 2n = 20 chromosomes. At the end of mitosis, each daughter cell would have ______ chromosomes
kozerog [31]

Answer:

At the end of mitosis, 2n = 20

At the end of meiosis I, n = 10

At the end of meiosis II, n = 10

Explanation:

Mitosis is a type of cell division in which daughter cell produced are genetically identical to their mother cell. So, no. of chromosome does not change after mitosis.

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

Meiosis is a type of cell division in which mother cell produces two haploid cells ones with a single set of chromosomes.

Meiosis is a two step cell division, Meiosis I and Meiosis II.

In meiosis I, homologous pair separates, so no. of chromosomes becomes half.

In meiosis II, sister chromatids separates. So, the number of chromosomes remains same (i.e. Have same no. of chromosome as present in cell produced after meiosis I).

So, at the end of mitosis, each daughter cell would have <u>20</u> chromosome.

At the end of meiosis I, each daughter cell would have n = 10 chromosomes. At the end of meiosis II, each daughter cell would have n = 10 chromosomes.

7 0
3 years ago
Which of these half-reactions represents reduction?
gogolik [260]

Answer: The half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element  takes place is called reduction-half reaction.

For example, the oxidation state of Cr in Cr_{2}O^{2-}_{7} is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}

Similarly, oxidation state of Mn in MnO^{-}_{4} is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

MnO^{2-}_{4} + 5 e^{-} \rightarrow Mn^{2+}

Thus, we can conclude that half-reactions represents reduction are as follows.

  • Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}
  • MnO^{-}_{4} \rightarrow Mn^{2+}
3 0
3 years ago
How can you balance this chemical equation?HSiCl3+H2O~H10Si10O15+HCL​
qwelly [4]
10HSiCl3 + 15H2O = H10Si10O15 + 30HCl
4 0
3 years ago
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