All categories

2 years ago

Help me out here please?

Chemistry

1 answer:

Tasya [4]2 years ago

3 0

Answer:

Explanation:

You might be interested in

Can anyone help me please this is due and i need help someone check over my answers

Yuliya22 [10]

4.) D
10.) C
12.) D
13.) D
14.) D
15.) D

5 0

3 years ago

What are the rows of the periodic table called?

Sergio039 [100]

Answer:

"The elements are arranged in seven horizontal rows,called periods or series,and 18 vertical columns,called Groups."

Explanation:

Risk*

4 0

2 years ago

A 200. gram sample of a salt solution contains 0.050 grams of NaCl. What is the concentration of the

wariber [46]

Answer:

2.5 × 10² ppm

Explanation:

Step 1: Given data

Mass of NaCl: 0.050 g

Mass of the sample: 200. g

Step 2: Convert 0.050 g to μg

We will use the conversion factor 1 g = 10⁶ μg.

0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg

Step 3: Calculate the concentration of NaCl in ppm

The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.

5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm

3 0

3 years ago

Round each number to four significant figures

ira [324]

Answer:

84,791 » 8479

256.75 » 256.8

431,801 » 4318

0.00078100 » 0.0007810

Explanation:

$.$

5 0

2 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago