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2 years ago

How does the freezing point of water from the ocean compare to distilled water one buys at the store?

Chemistry

2 answers:

Lelechka [254]2 years ago

8 0

Due to prescence of any impurity, there will be change in physical properties of any liquid.

sdas [7]2 years ago

5 0

The answer is
The ocean water will have a lower freezing point.

distilled water is 0 °C
ocean water is -1.33°C

Because the solute in the sea water decrease the freezing point

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One mole of acetyl chloride was mixed with one mole of dimethylamine. After the reaction is complete, what species can be found

fredd [130]

Answer:

N,N-dimethylacetamide is formed.

Explanation:

It is an example of a nucleophilic addition-elimination reaction. Here dimethylamine acts as a nucleophile.
In the first step, dimethyl amine gives nucleophilic addition reaction at carbonyl center of acetyl chloride.
In the second step, removal of Cl atoms occurs.
In the third step, deprotonation takes place from amino group to produce N,N-dimethylacetamide.
Full reaction mechanism has been shown below.

7 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

A radioactive isotope of potassium (K) has a half-life of 20 minutes. If a 40.0 gram sample of this isotope is allowed to decay

Ivahew [28]

Answer: 2.5 grams

Explanation:

5 0

2 years ago

Read 2 more answers

How many moles of oxygen are required in order to produce 4 moles of<br> water?

Cloud [144]

16 I think hopefully

4 0

2 years ago

How many molecules of H2O are there in 1.0 g of H2O?

NISA [10]

Molar mass H₂O = 18.0 g/mol

number of moles :

1.0 / 18.0 => 0.055 moles

1 mole -------------- 6.02 x 10²³ molecules
0.055 moles -------- ? molecules

molecules = 0.055 x ( 6.02 x 10²³) / 1

molecules = 3.311x10²² / 1

= 3.311 x 10²² molecules

hope this helps!

8 0

2 years ago