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Andrews [41]
2 years ago
8

What are the products for the following equation?NaCl + H2O ->​

Chemistry
1 answer:
krek1111 [17]2 years ago
3 0

Answer:

NaOH + HCI

Explanation:

yesssssssssssss

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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
Which of the following is NOT true about equilibrium?
podryga [215]

Answer:

c

Explanation:

6 0
3 years ago
PLEASE HELPP<br> Tell me everything you know about "balancing the equation" for science
Free_Kalibri [48]
Watch melissa maribel explains it amazingly on her yt channel
3 0
3 years ago
The melting point of table salt is 801 degrees celsius. is this a physical or chemical property?
Valentin [98]
It is a Physical property because like water it changes into a ice cube but it can be melted into water or be turned back into an ice cube.
5 0
3 years ago
To what volume should you dilute 50.0 ml of 12 m hno3 solution to obtain a 0.100 m hno3 solution?
bearhunter [10]

Answer:

The answer is "6L"

Explanation:

Formula:

\bold{C_1 \times V_1 = C_2 \times V_2 }\\\\V_2= \frac{C_1\times V_1}{C_2}

Values:

\to C_1= 12 \ m\\\to V_1= 50 \ ml\\\to C_2= 0.100 \ m\\\\\\V_2= \frac{12 \times 50 }{0.100}

   = \frac{12 \times 50 }{0.100}\\\\= \frac{12 \times 50 \times 1000}{100}\\\\= \frac{600 \times 1000}{100}\\\\= 600 \times 10\\\\=6000 \ ml\\= 6 \ L

4 0
3 years ago
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