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3 years ago

What is the element with the symbol Ba?

Chemistry

2 answers:

liubo4ka [24]3 years ago

8 0

Answer:

Barium is a chemical element with the symbol Ba and atomic number 56. It is the fifth element in group 2 and is a soft, silvery alkaline earth metal.

Explanation:

Vika [28.1K]3 years ago

3 0

Answer:

Barium

Explanation:

-Atomic #56

-5th element inn group 2

- soft

- silvery alkaline earth metal

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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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3 years ago

Swimming pool structures and structural _____ [680.26(B)(1) and (B)(2)] shall not be used as a grounding electrode.

leonid [27]

A grounding electrode is any object that directly links to the earth. They are most times used to divert electricity from the elements.

Swimming pool structures and structural <u>reinforcing steel. 250.52(B)(3)</u><u>,</u> [680.26(B)(1), and (B)(2)] shall not be used as a grounding electrode.

In code 250.52(B)(3) it is clearly specified that the bonding grid and reinforcing steel that is related to a pool should not be used as grounding electrodes.

This is essential because when a metal that lies beneath a swimming pool is used as a grounding electrode, current from nearby electrical systems can be introduced into the pool.

This could cause the electrocution of anybody in the swimming pool at that time.

Learn more here:

brainly.com/question/14681208

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2 years ago

Determine the mass of SO₂ that contains 6.075 × 10^26 S atoms.

Misha Larkins [42]

Avogadro's law states that in a mole of any substance, there are $6.022 \times 10^{23}$ atoms. This means that in the given sample, there are