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ASHA 777 [7]
3 years ago
14

HELP PLZ

Chemistry
1 answer:
Vinvika [58]3 years ago
8 0

Answer:

1). 19.9

2). 0

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Select the correct answer.
ankoles [38]

It would still have oceans but no atmospheric water in Earth if no icy debris had arrived.

A.  It would still have oceans but no atmospheric water.

<u>Explanation:</u>

Seas characterize our home planet, covering most of the Earth's surface and driving the water cycle that commands our territory and climate. However, progressively significant still, the narrative of our seas wraps our home in a far bigger setting that ventures profound into the universe and spots us in a rich group of sea universes that range our nearby planetary group and past.

It would in any case have seas yet no air water on Earth if no frigid flotsam and jetsam had shown up. For a long time, it was accepted that the frosty moons were only that - solidified husks, strong to their center. However, lately that thought has steadily been supplanted by a fresher, additionally energizing worldview.

4 0
3 years ago
A waste treatment pond is 50 m long and 25 m wide, and has an average depth of 2 m. The density of the waste is 75.3lbm/ft^3 .Ca
umka21 [38]

Answer:

W = 6.65 \cdot 10^{6} lbf

Explanation:

To find the weight (W) of the pond contents first we need to use the following equation:

W = m\cdot g   (1)

Where m the mass and g is the gravity  

Also, we have that the mass is:

m = \rho*V  (2)    

Where ρ is the density and V the volume

We cand calculate the volume as follows:

V = L*w*d   (3)

Where L is the length, w is the wide and d is the depth  

By entering equation (2) and (3) into (1) we have:

W = \rho*L*w*d*g

W = 75.3 lbm/ft^{3}*50 m*25 m*2 m*9.81 m/s^{2}  

W = 75.3 lbm/ft^{3}*\frac{(1 ft)^{3}}{(0.3048 m)^{3}}*\frac{0.454 kg}{1 lbm}*50 m*25 m*2 m*9.81 m/s^{2} = 2.96 \cdot 10^{7} N}*\frac{0.2248 lbf}{1 N} = 6.65\cdot 10^{6} lbf              

Therefore, the weight of the pond is 6.65x10⁶ lbf.

I hope it helps you!

6 0
3 years ago
A 0.300 m solution of hcl is prepared by adding some 1.50 m hcl to a 500 ml volumetric flask and diluting to the mark with deion
liberstina [14]
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
8 0
3 years ago
How many protons z and how many neutrons n are there in a nucleus of the most common isotope of silicon, 2814si?
zheka24 [161]
The 28 is the sum of the protons and neutrons in the element silicon.
ALL silicon atoms have 14 protons in the nucleus, so we can turn this into an equation:
#protons + #neutrons = 28
14 + #neutrons = 28
#neutrons = 14
#protons = 14
7 0
3 years ago
Read 2 more answers
What is the frequency of a photon that has a wavelength of 6.58 x 10-8m?
julsineya [31]

Answer:

6 x 10 (power to the 15) H2

5 0
3 years ago
Read 2 more answers
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