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Advocard [28]
2 years ago
14

Which subshell holds the valence electrons of barium?.

Chemistry
1 answer:
olga2289 [7]2 years ago
7 0

Answer:

6s

Explanation:

Barium is in group 2 of the s block and is in period 6.

You might be interested in
9
Svetllana [295]

Answer:

0.0025moles

Explanation:

Molarity of a solution (M) = number of moles (n) ÷ volume (V)

According to this question, to make 250 mL of a 0.01 M solution of CaCl, the following number of moles is needed:

Volume = 250mL = 250/1000 = 0.250Litres.

Using; molarity = n/V

0.01 = n/0.250

n = 0.0025

n = 2.5 × 10^-3 moles.

5 0
3 years ago
Is the answer B? Help
Masja [62]

Answer:

A

Explanation:

Hmm, so we have the following in the diagram

Pt(s)

Cl2(g)

Ag(s)

NaCl(aq)

AgNO3(aq)

Pt 2+, 4+, 6+  Though it states Pt is inert

Cl 2-

Ag 1+

Na 1+

NO3-

Anode definition: the positively charged electrode by which the electrons leave an electrical device.

Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.

Cations attracted to cathode pick up electrons

Anions attracted to anode release electrodes+

Reduction at Cathode (red cat gain of e)

Oxidation at Anode (ox anode loss of e)

So from the diagram we can see that the charge is being generated through the 2 metal plates.

So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-

7 0
3 years ago
a compound is 21.20% nitrogen,6.06% hydrogen, 24,30% sulfur, and 48.45% oxygen. write the empirical formula and name the compoun
IRISSAK [1]

The numbers may be off by a few, if you worked it out, since my periodic table is for AP chemistry. Hope this helps!!!

4 0
4 years ago
86.14 mL of an acid solution was needed to neutralize 30.24 mL of a base solution of unknown concentrations. A second trial is r
Airida [17]

Answer:

The correct answer is option B.

Explanation:

As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.

After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.

5 0
3 years ago
Nitric acid can be produced by the reaction of gaseous nitrogen dioxide with water. 3 no2(g + h2o(? ?? 2 hno3(? + no(g if 538 l
Phantasy [73]
The balanced chemical reaction is written as:

<span>3NO2 + H2O = 2HNO3 + NO

Assuming that the gases in this reaction are ideal gas, then we can use the conversion from L to moles which is 1 mol of ideal gas is equal to 22.4 L. We calculate as follows:

538 L NO2 ( 1 mol / 22.4L ) ( 1 mol NO / 3 mol NO2 ) ( 22.4 L / 1 mol ) = 179.33 L NO is produced</span>
8 0
3 years ago
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