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kondaur [170]
3 years ago
13

HELP MEE

Chemistry
2 answers:
aniked [119]3 years ago
6 0

Answer:

Cedar, mahogany, red wood or something else is fine, but the only downside is that it will cost you more. However, a project that is partially or fully exposed to the elements, something more than normal lumber is a must. The most cost effective method is to use pressure treated lumber.

Explanation:

Murrr4er [49]3 years ago
3 0

Answer:

What he/she said ↑↑↑↑

Explanation:

This is because if you play in the summer it becomes very hot, and your body would burn

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
Moraines left by glaciers are different from deposits left by rivers because the rocks left behind are _____.
12345 [234]

Explanation:

unorganized and unsorted i think this is the answer

7 0
3 years ago
When the equation Eg(s) + DnGO4(aq) ---> Dn(s) + Eg2(GO4)3(aq) is balanced, what is the coefficient in
Artemon [7]

B) 2

You would first balance GO4 by adding a coefficient of 3 in front of DnGO4 in the reactants. Then you’d balance the 3DnGO4 by adding a coefficient of 3 in front of Dn in the products. Finally you’d balance Eg by adding a coefficient of 2 in front of Eg to balance with the Eg2(GO4)3 in the products.
5 0
3 years ago
Which of the following can form a hydrogen bond with the HF molecule?
Lostsunrise [7]

Answer:

helium

(He)

Explanation:

helium

(He)

6 0
2 years ago
Calculating Density Warm Up
Artist 52 [7]

Answer:

2

Explanation:

33-25=8

48/8=6

6 0
3 years ago
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