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Fittoniya [83]
2 years ago
15

During a chemical reaction ____ is never created or destroyed

Chemistry
1 answer:
arsen [322]2 years ago
7 0

Answer:

Matter

Explanation:

The Law of Conservation of Mass

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Please help on 44. /45./ 46./47,
Aleonysh [2.5K]
The only one that I can do without google is 47. Sorry that I can't answer the others. The answer to 47 is this: you know that the western side of the hill has the steepest slope because the ovals showing altitude are way closer together. The closer the circles/ovals are, the steeper the slope is.

Sorry if this doesn't help much, but I answered what I could without cheating.

Foxeslair
4 0
3 years ago
The atomic mass is equal to what?​
dsp73

Answer:

The mass number

Explanation:

I hope this helps.

4 0
3 years ago
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What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D
bekas [8.4K]

Answer: D) 2.38*10^-^2^2 g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})

= 2.38*10^-^2^2 g

So, the correct option is D) 2.38*10^-^2^2 g

7 0
3 years ago
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Use the periodic table to write the electron configuration for rubidium (Rb) in noble has notation
Alexxx [7]

Answer:

Rb: [Kr] 5s  

Step-by-step explanation:

Rb is element 37, the first element in Period 5.

It has one valence electron, so its valence electron configuration is 5s.

The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.

The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.

4 0
3 years ago
Please help me with this chemistry problem
Margarita [4]

Answer:

50 g of K₂CO₃ are needed

Explanation:

How many grams of K₂CO₃ are needed to make 500 g of a 10% m/m solution?

We analyse data:

500 g is the mass of the solution we want

10% m/m is a sort of concentration,  in this case means that 10 g of solute (K₂CO₃) are contained in 100 g of solution

Therefore we can solve this, by a rule of three:

In 100 g of solution we have 10 g of K₂CO₃

In 500 g of solution we may have, (500 . 10) / 100 = 50 g of K₂CO₃

6 0
3 years ago
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