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mojhsa [17]
3 years ago
15

Objects that can bew seen only under magnification

Chemistry
1 answer:
timofeeve [1]3 years ago
4 0
The answer is (D) microscopic. You can remember this, because the name is very close to "microscope," an instrument used to greatly magnify and observe tiny organisms and objects.
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Which sequence of group 18 elements demonstrates a gradual decrease in forces of attraction
DanielleElmas [232]

Answer: Xe, Kr, Ar, Ne

Explanation:

Hope i helped please mark brainliest!

8 0
3 years ago
If you are measuring something with a graduated cylinder, what are you measuring?
JulijaS [17]
With a graduated cylinder, you are most likely measuring mL (milliliters) or volume of a liquid, or the mass of a liquid.
8 0
3 years ago
A flask contains 5.0g of neon gas at a temperature of 35°C. The pressure gauge indicates 0.37atm inside the flask. What is the v
DaniilM [7]

Answer:

The volume of the neon gas is 17.07 L.

Explanation:

The ideal gas equation describes the relationship among the four variables  P, V, T, and n. An ideal gas<u> is a hypothetical gas whose pressure-volume-temperature  behavior can be completely accounted for by the ideal gas equation</u>.

In order to calculate the volume, first we need to convert the grams of neon to moles:

28.18 g Ne ------ 1 mol Ne

5.0 g Ne---------- <u>x= 0.25 mol Ne</u>

Now, using the ideal gas equation we calculate the volume:

pV= nRT

V= nRT ÷ p

V= (0.25 mol × 0.082 L.atm/mol.K × 308.15 K ÷ 0.37 atm

V= 17.07 L

6 0
3 years ago
What is the pH of a solution in which [OH-] = 1.0 x 10-2 M
Degger [83]

Answer:

12

Explanation:

pOH = -log(1.0 x 10^-2) = 2

pH = 14 - pOH

pH = 14 - 2 = 12

7 0
3 years ago
A food processing plant discharges 40 cfs (cubic feet per second) of process water containing an ultimate BOD (L0) of 25 mg/L an
Feliz [49]

Answer:

The right solution according to the question is provided below.

Explanation:

According to the question,

(a)

The initial conditions will be:

DO = \frac{(40\times 1.8)+(260\times 7.6)}{40+260}

      = \frac{2048}{300}

      = 6.826 \ mg/L

The initial oxygen defict will be:

Do = 8.5-6.826

     = 1.674 \ mg/L

The initial BOD will be:

Lo = \frac{(40\times 25)+(260\times 3.6)}{40+260}

    = \frac{1936}{300}

    = 6.453 \ mg/L

(b)

The time reach minimum DO:

tc = \frac{1}{(kr-kd)} ln{(\frac{0.76}{0.61} )[1-\frac{1.674(0.76-0.61)}{0.61\times 6.453} ]}

   = \frac{1}{0.15}\times ln \ 1.158

By putting the values of log, we get

   = 0.973 \ days

The distance to reach minimum DO will be:

Xc = \frac{1.1\times 3600\times 24}{5280}\times 0.973 \ days

    = 18\times 0.973

    = 17.5 \ miles

8 0
3 years ago
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