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3 years ago

Is Columns of elements in the periodic table are called periods correct in the underlined sentence

Chemistry

1 answer:

SSSSS [86.1K]3 years ago

6 0

I don't know sorry I really need point

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In the process of making soap, I poured some of the cooked mixture through some muslin fabric, in order to separate the solid pa

lyudmila [28]

Answer:

filtering

Explanation:

you're pouring the mixture through muslin cloth to keep the particles and bigger peaces out of the soap.

6 0

3 years ago

Why is mercury the only metal to have been used in thermometers

MArishka [77]

I think because its the only one to be liquid at normal temperatures.

6 0

4 years ago

15 points!

Alex_Xolod [135]

Answer:

C: It is the difference between the total kinetic and potential energy in a system.

Explanation:

6 0

3 years ago

Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov

love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

$V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3$

Mass of the Astroturf = m

Density of the Astroturf = d = 187 $oz/ft^3$

$d=\frac{m}{V}$

$m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz$

1 oz = 0.0283495 kg

$m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg$

Weight of the Astroturf = W

W = mg

$=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N$

The weight of the Astroturf is 179,684.31 Newtons.

8 0

3 years ago

What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?

ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M $H_{3}PO_{4}$ from 100 ml of 0.5 M solution.

Explanation:

Given: $M_{1}$ = 0.1 M, $V_{1}$ = ?

$M_{2}$ = 0.5 M, $V_{2}$ = 100 mL

Formula used to calculate the volume of water is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL$

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M $H_{3}PO_{4}$ from 100 ml of 0.5 M solution.

7 0

3 years ago