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Ann [662]
4 years ago
12

What is the rule of Conservation of Mass?

Chemistry
2 answers:
yan [13]4 years ago
8 0
The mass of the products in a chemical reaction must equal the mass of the reactants.
allsm [11]4 years ago
4 0

Answer: It is the principle that matter cannot be created or destroyed.

Explanation:

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6 0
4 years ago
What volume of a 0.424 M CO2 solution are needed to produce 63.58 mL of 0.731 M solution of K2CO3?
cricket20 [7]

Answer:

Socratic app

Explanation:

it will help you

6 0
3 years ago
If 0.500 L of a 4.00-M solution of copper nitrate, Cu(NO3)2, is diluted to a volume of 1.50 L by the addition of water, what is
Wewaii [24]

Answer:

1.33 M

Explanation:

We'll begin by writing out the data obtained from the question. This includes the following:

Volume of the stock solution (V1) = 0.5L

Molarity of the stock solution (M1) = 4M

Volume of diluted solution (V2) = 1.5L

Molarity of the diluted solution (M2) =.?

With the application of the dilution formula, the molarity of the diluted solution can be obtained as follow:

M1V1 = M2V2

4 x 0.5 = M2 x 1.5

Divide both side by 1.5

M2 = (4 x 0.5) / 1.5

M2 = 1.33 M

Therefore the molarity of the diluted solution is 1.33 M

8 0
3 years ago
Calculate the molecular mass of Co(NO3)2<br>with steps please
Nana76 [90]

Answer:

182.933~ \text{g mol}^{-1}

Explanation:

\text{Atomic mass of cobalt(Co)} = 58.933~u\\\\\text{Atomic mass of nitrogen(N)}= 14~u\\\\\text{Atomic mass of oxygen(O)}=16~u\\\\\text{Molar mass of}~ \text{Co}\left(\text{NO}_3} \right)_2 = 58.933 + 2(14 +16 \times 3)~ \text{g mol}^{-1}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(58.933 + 124)~ \text{g mol}^{-1}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=182.94~ \text{g mol}^{-1}

4 0
2 years ago
The specific heats and densities of several materials are given below: Material Specific Heat (cal/g·°C) Density (g/cm3) Brick 0
abruzzese [7]

<u>Answer:</u> The change in temperature is 84.7°C

<u>Explanation:</u>

To calculate the change in temperature, we use the equation:

q=mc\Delta T

where,

q = heat absorbed = 1 kCal = 1000 Cal    (Conversion factor: 1 kCal = 1000 Cal)

m = mass of steel = 100 g

c = specific heat capacity of steel = 0.118 Cal/g.°C

\Delta T = change in temperature = ?

Putting values in above equation, we get:

1000cal=100g\times 0.118cal/g^oC\times \Delta T\\\\\Delta T=\frac{1000cal}{100g\times 0.118cal/g^oC}\\\\\Delta T=84.7^oC

Hence, the change in temperature is 84.7°C

5 0
3 years ago
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