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dimulka [17.4K]
3 years ago
8

An engineer that designs buildings is a (n)

Chemistry
1 answer:
guapka [62]3 years ago
7 0
An architectural engineer.<span />
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The temperature of the water was measured at different depths of a pond.
rusak2 [61]
A. Dependent variable
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3 years ago
If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanog
9966 [12]

Answer:

616,0 ng is the right answer.

Explanation:

You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

1 .10^9 nanomoles ...................... 56.0 gr

11 nanomoles .....................

(11 x 56) / 1 .10^9 nanomoles = 6.16 x 10^-7 gr

Let's convert

6.16 x 10^-7 gr x 1 .10^9 = 616 ngr

8 0
3 years ago
The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium
PilotLPTM [1.2K]

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

 \frac{9 .84x10^8}{1000000}      = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = \frac{mass}{volume}

mass = density x volume

mass = 1 x 984

         = 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

4 0
3 years ago
Calculate the change in entropy when 1.00 kg of water at 100 ∘C is vaporized and converted to steam at 100 ∘C. Assume that the h
andrew11 [14]

Answer : The change in entropy is 6.05\times 10^3J/K

Explanation :

Formula used :

\Delta S=\frac{m\times L_v}{T}

where,

\Delta S = change in entropy = ?

m = mass of water = 1.00 kg

L_v = heat of vaporization of water = 2256\times 10^3J/kg

T = temperature = 100^oC=273+100=373K

Now put all the given values in the above formula, we get:

\Delta S=\frac{(1.00kg)\times (2256\times 10^3J/kg)}{373K}

\Delta S=6048.25J/K=6.05\times 10^3J/K

Therefore, the change in entropy is 6.05\times 10^3J/K

5 0
3 years ago
What type of pollution in cities composed of car exhaust and industrial pollutes
Troyanec [42]
Air pollution is the answer

4 0
3 years ago
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