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Vaselesa [24]
3 years ago
15

1.20×10−8s to nanoseconds

Chemistry
1 answer:
AveGali [126]3 years ago
7 0

Answer:

There are 12 nanoseconds in 1.2\times 10^{-8}\ s.

Explanation:

We need to convert 1.2\times 10^{-8}\ s to nanoseconds.

We know that,

1\ s=10^9\ ns

Now using unitary method to solve it such that,

1.2\times 10^{-8}\ s=1.2\times 10^{-8}\ \times 10^9\\\\=1.2\times 10\\\\=12\ ns

So, there are 12 nanoseconds in 1.2\times 10^{-8}\ s.

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A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
What type of weathering breaks down into smaller pieces
kodGreya [7K]

Answer:Mechanical weathering

weathering breaks down rocks by forming new minerals that are stable at the Earth's surface.

8 0
3 years ago
How does carbon exist in the atmosphere?
maks197457 [2]

Answer:

Carbon is found in the atmosphere mostly as carbon dioxide. Animal and plant respiration place carbon into the atmosphere. When you exhale, you are placing carbon dioxide into the atmosphere. Carbon is found in the lithosphere in the form of carbonate rocks. Carbonate rocks came from ancient marine plankton that sunk to the bottom of the ocean

Explanation:

hope this helps

6 0
3 years ago
What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?
gregori [183]

The ideal gas under STP is 22.4 L/mol. While the gas has a rule of P1V1/T1=P2V2/T2. So the volume under 101 kPa and 273 K is 0.2*22.4=4.48 L.


7 0
3 years ago
Read 2 more answers
A 45.0 g sample of a metal at 85.6 °C is placed in 150.0 g of water at 24.6 °C. The final temperature of the system is 28.3 °
earnstyle [38]

Answer:

904.014 j/kgk

Explanation:

Mass of metal = 45g

Temperature of metal = 85.6°

Mass of water = 150

Temperature of water = 24.6

Final temperature of system = 28.3

Heat lost by metal = Heat gained by water

m1 * c1 * dt = m2 * c2 * dt

Q = quantity of heat

Q = m*c*dt

dt = change in temperature

dt of water = 28.3 - 24.6 = 3.7

dt of metal = 85.6 - 28.3 = 57.3

Specific heat capacity of water, c = 4200

(45 * 10^-3) * c * 57.3 = (150 * 10^-3) * 4200 * 3.7

2.5785c1 = 2331

c1 = 2331 / 2.5785

= 904.01396

= 904.014 j/kgk

3 0
2 years ago
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