All categories

3 years ago

__________ may be shared by or transferred to other atoms.

Chemistry

2 answers:

Vsevolod [243]3 years ago

8 0

D.) "Electrons" may be shared by or transferred to other atoms.

Hope this helps!

Lisa [10]3 years ago

8 0

Electrons can and are often shared with other atoms.

The elemental group that gives their electrons up the easiest are the Alkali Metals, while the group that holds the tightest is the Halogen group.

You might be interested in

4. which answer below is the formula mass of a water molecule?

mars1129 [50]

The answer is 18.02 g/mol

6 0

3 years ago

What is the correct ionic formula when mg2+ and p3− react?

Mumz [18]

Answer: Ionic formula will be $Mg_3P_2$ .

Explanation: $Mg^{2+}$ and $P^{3-}$ ions will form a ionic compound. Ionic compounds have both metals and non-metals.

Here $Mg^{2+}$ is a metal and $P^{3-}$ is a non-metal.

The net charge on any compound must be 0.

So we need 2 phosphate ions to balance the charge on $Mg^{2+}$ ions. Similarly we need 3 Magnesium ions to balance the charge on $P^{3-}$ ions.

Criss-crossing the charges, we will get the formula as $Mg_3P_2$

Criss-crossing is shown in the image below.

6 0

3 years ago

Read 2 more answers

Which of the following molecules has correctly labeled bond dipoles?

Pepsi [2]

The bond dipole moment uses the idea of electric dipole moment to measure the polarity of a chemical bond within a molecule. It occurs whenever there is a separation of positive and negative charges. In the diagram above, option B exhibited a bond dipole moment. I hope this helps.

8 0

2 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

7 0

3 years ago

Which of the following tools is not used for measurement?

Verdich [7]

Answer:

tape measure

Explanation:

I am helping

8 0

2 years ago