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Fynjy0 [20]
3 years ago
8

A variable is something that can never be changed in an experiment. False True

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0
I’m pretty sure it true
You might be interested in
What is the density of a block of marble that occupies 277 cm3 and has a mass of 928 g? Answer in units of g/cm3 .
Evgesh-ka [11]
V=277cm^{3}\\
m=928g\\\\
d=\frac{m}{V}=\frac{928g}{277cm^{3}}\approx3,35\frac{g}{cm^{3}}
8 0
3 years ago
How many particles are present in 8.0 moles of silver?
Zielflug [23.3K]

Answer:the answer is b

Explanation:I took the test and got it right

3 0
3 years ago
Determine the number of valence electrons for the following: [kr] 5s2 4d6
wlad13 [49]

Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).

Explanation:

This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7

Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4

3 0
3 years ago
Can someone help me with problem 21.105? Many thanks!
Reika [66]
The correct answer is 1 to the 3rd power
3 0
2 years ago
Consider the reaction 2 S + 3 O2 → 2 SO3 , which has a 75.1% yield. How much O2 is consumed if 583 g of SO3 are produced?
Ymorist [56]

Answer:

A/1.      10.9 mol O2

Explanation:

583 g x 1 mol SO3 x 3 mol O2 /

      80.057 g mol SO3 x 2 mol SO3

- You just need to find molar mass of SO3, which is 80.057 g.

- Everything else came from formula. Further explanation...

- Always start with what they give, such as 583 g. Then find 1 mol of what is being produced, in this it is SO3. We already found this because we did molar mass above. Next. find how many moles of what they want, which is O2. Look in equation and you can see 3 mol in from of O2. Next, do the same for SO3 and you can find 3 mol in front of that. Lastly, just do the math.

- If you need a further explanation or more help on any problems I would be happy to help, just let me know.

4 0
3 years ago
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