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mars1129 [50]
3 years ago
5

What is the percent yield if 155 grams of calcium carbonate is treated with 250 grams of hydrichloric acid andb142 grams of calc

iym chloride is obtained
Chemistry
2 answers:
dlinn [17]3 years ago
7 0

Answer:

375 grams

Explanation:

RideAnS [48]3 years ago
5 0

Answer:

375

Explanation:

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You throw a 2-kg rock with a force of 10 N. What is the acceleration of the rock?
Tema [17]

Answer:

<h2>5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{10}{2}  \\

We have the final answer as

<h3>5 m/s²</h3>

Hope this helps you

6 0
3 years ago
Section Review<br> 22<br> 1. What is meant by the term continuous spectrum?
just olya [345]
A set of attainable values for some physical quantity such as energy or wavelength
5 0
3 years ago
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
How many particles are in one mole of copper (II) sulfate, CuSO4?
Kobotan [32]
2. Avogadro’s number
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3 years ago
Please help!
cricket20 [7]
1- metal and non metal
2- true
3- chlorine
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3 years ago
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