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charle [14.2K]
3 years ago
14

A good team member will actively listen, which means they will ______.

Chemistry
2 answers:
Wittaler [7]3 years ago
8 0
B is the answer I think better sure
Sergeeva-Olga [200]3 years ago
8 0

Answer:

C. use body language to understand how a speaker feels

Explanation:

let me know if i'm right

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I need help with this question. Thank you.
marishachu [46]
There are 76 atoms in total
3 0
3 years ago
Which of these solutes raises the boiling point of water the most?
Vlad [161]
I think the answer would be Ionic sodium phosphate (Na3PO4) because it has the greatest boiling point elevation.
7 0
3 years ago
Read 2 more answers
13. Fill in the following table
Mandarinka [93]

Answer:

Potassium Bromide = KBr

Nitrogen dioxide = NO₂

Lithium oxide = Li₂O

Explanation:

Potassium Bromide (KBr)

4 0
3 years ago
A field worker is exposed to a xylene for a duration of 8 weeks at 40 hrs/wk. The concentration of xylene in the workplace is 40
Andrej [43]

Answer:

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

Explanation:

Number of hours worker exposed to xylene = 40 hr/week\times 8 week = 320 hours

The concentration of xylene in the workplace =40 \mu g/m^3

The worker is inhaling air at a rate of 0.9 m^3/hr.

Amount xylene inhaled by worker in an hour :

= 40\mu g/m^3\times 0.9 m^3/hr=36 \mu g/hr

Amount xylene inhaled by worker in 320 hours:

36 \mu g/hr\times 320 hr=11,520 \mu g=11,520\times 0.001 mg=11.520 mg

1 μg = 0.001 mg

Amount xylene inhaled by worker in 320 hours = 11.520 mg

1 day = 24 hours

Amount xylene inhaled by worker in 1 day:

\frac{24}{320}\times 11.520 mg=0.864 mg

Assuming 70 kg body mass, the chronic daily intake of xylene :

\frac{0.864 mg/day}{70 kg}=0.01234 mg/ kg day\approx 0.012 mg/ kg day

The chronic daily intake during the period of exposure is most nearly 0.012 mg/kg day.

5 0
3 years ago
What mass (in grams) of silver contains the same number of atoms as 5.59 grams of sulfur?
Amanda [17]

Answer:

18.84 g of silver.

Explanation:

We'll begin by calculating the number atoms present in 5.59 g of sulphur. This can be obtained as follow:

From Avogadro's hypothesis,

1 mole of sulphur contains 6.02×10²³ atoms.

1 mole of sulphur = 32 g

Thus,

32 g of sulphur contains 6.02×10²³ atoms.

Therefore, 5.59 g of sulphur will contain = (5.59 × 6.02×10²³) / 32 = 1.05×10²³ atoms.

From the calculations made above, 5.59 g of sulphur contains 1.05×10²³ atoms.

Finally, we shall determine the mass of silver that contains 1.05×10²³ atoms.

This is illustrated below:

1 mole of silver = 6.02×10²³ atoms.

1 mole of silver = 108 g

108 g of silver contains 6.02×10²³ atoms.

Therefore, Xg of silver will contain 1.05×10²³ atoms i.e

Xg of silver = (108 × 1.05×10²³)/6.02×10²³

Xg of silver = 18.84 g

Thus, 18.84 g of silver contains the same number of atoms (i.e 1.05×10²³ atoms) as 5.59 g of sulfur

7 0
2 years ago
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