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3 years ago

1 point

Chemistry

1 answer:

Genrish500 [490]3 years ago

5 0

With an electronegativity of 0.89, Barium requires the least amount of energy to remove its valence electrons.

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2CO + O2  2CO2

murzikaleks [220]

Answer:

75 L Co2

37.5 L o2

Explanation:

welcome.

3 0

2 years ago

Write the electron configurations of cu2+ and ag+. is each electron configuration consistent with the color you observed for eac

djverab [1.8K]

(1) Copper:
Copper has the atomic number : 29
Cu2+ means that it lost 2 electrons, therefore, the total number of electrons is 29-2 = 27 electrons.
The electronic configuration is:
1s2 2s2 2p6 3s2 3p6 3d9

(2) Silver:
Silver has atomic number : 47
Ag + means that it lost one electron, therefore, the total number of electrons is 47-1 = 46 electrons
The electronic configuration is:
<span>1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10</span>

8 0

3 years ago

Read 2 more answers

What mass would release 2620J of heat when cooled 50 degrees C?

OverLord2011 [107]

the answer is B)12.5

7 0

3 years ago

Read 2 more answers

Which disease is caused when white blood cells are produced in excess?

marin [14]

Answer:

it can cause blood cancer (leukemia)

4 0

3 years ago

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

3 years ago