Non-metal ions are negative because...

Please help me with this question it’s due in five minutes I’ll give brainliest

zysi [14]

I believe the correct answer is A) 6

5 0

3 years ago

You would be most likely to use a slicing machine if you were using the __________ method to produce cookies.

Verizon [17]

You would be most likely to use a slicing machine if you were using the <u>icebox </u>method to produce cookies.

In the icebox method a type of cookie in which the dough is made, rolled into a stick, and refrigerated until the dough hardens. The dough can be removed from the refrigerator, cut into individual pieces, and then baked. The rest of the dough is returned to the refrigerator until needed.

Icebox method, also known as refrigerator cookies, are sliced and baked cookies. The dough is formed into logs, chilled in the refrigerator (also called an icebox), sliced , and then baked.

Learn more about the Icebox method here,

https://brainly.in/question/1513677

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7 0

1 year ago

What does the group number of element 114 have to do with its stability?

zhenek [66]

Answer:

Super-heavy elements like 114 usually only exist for fractions of a second. ... The physicists called these magic numbers the “island of stability”, because the elements with the numbers cluster together on the periodic table, flanked on all sides by ephemeral elements that dissipate in nanoseconds.

8 0

3 years ago

Name the following alkyne.<br> Please help me <3

Alexandra [31]

Answer:

D. 7-methyl-3-octyne

Explanation:

1) Identify the parent chain and name it like an alkane.

• The longest chain of carbons, which consists of the functional group (which is the alkyne group in this case: C≡C).

• There are 8 carbons in the longest chain, so it is called an octane.

2) Now, identify the location of the functional group.

• The location number of the functional group should be as low as possible. Thus although we could count from the right, we start counting from the left since the functional group is closer to the left. From the left, the functional group would be at carbon 8 while from the left, it is on carbon 3.

• Replace 'ane' with 'yne' in octane for the alkyne group.

• This would give us 3-octyne.

3) Lastly, add in the name of the branch.

• Here we have one branch, -CH₃. This is read as methyl.

• Identify the location number of the branch by counting the number of carbons in the same direction as when we counted the location number of the functional group. The methyl branch has a location number of 7.

• Adding the name of the branch before the parent chain, we would arrive at 7-methyl-3-octyne as the IUPAC name of the alkyne.

Further Explanation:

A) This option is incorrect as there are only 8 carbons in the parent chain. Although there are 9 carbons in total, the 9th carbon is taken care of in '7-methyl'.

B) Location number of the functional group should be as low as possible, so start counting the number of carbons from the left!

C) Since the functional group is an alkyne, the word 'octane' should be 'octyne' instead.

8 0

2 years ago

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$

Reduction half reaction (Cathode) : $Ag^{+}(aq)+e^-\rightarrow Ag(s)$

Thus the overall reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0

3 years ago