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2 years ago

Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:

Chemistry

1 answer:

Lapatulllka [165]2 years ago

4 0

Answer:

a) The element is Manganese (Mn)

b) The element is Zirconium (Zr)

Explanation:

The step by step analysis and explanation is as shown in the attachment

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1. Swimming pool manufacturers recommend maintaining the

Viefleur [7K]

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

"ppm" of "parts per million" is a unit of concentration equivalent to milligrams of solute per liters of solution.

A pool must maintain a chlorine concentration of 3.0 ppm (3.0 mg/L). The mass of chlorine in 3.4 × 10⁶ L is:

3.0 mg Cl₂/L × 3.4 × 10⁶ L = 1.0 × 10⁷ mg Cl₂

A 3.4 × 10⁶ L swimming pool must have a mass of 1.0 × 10⁷ mg Cl₂ to maintain a concentration of 3.0 ppm.

Learn more about ppm here: brainly.com/question/13395702

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2 years ago

What is one way a pathogen can enter the body?

Ipatiy [6.2K]

I’m pretty sure it’s Nose

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3 years ago

Read 2 more answers

Which is an element? A: H20 B: NaCl C: Mg D: CO

Talja [164]

The answer is option D "CO." Co also known as Cobalt is the 27th element on the periotic table. It was discovered in 1735, it's boiling point is 3200 k.

Atomic mass: 58.9332
Protons: 27
Neutrons: 32
Electrons: 27

Hope this helps!

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3 years ago

Read 2 more answers

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

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3 years ago

What happens to concrete when it's no longer needed

Sveta_85 [38]

Did you mean when it is a liquid?

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2 years ago