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Free_Kalibri [48]
3 years ago
7

2. List three types of chemical changes. Help me

Chemistry
1 answer:
Sedaia [141]3 years ago
6 0
Burning wood, rusting of iron, and baking a cake
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A certain object has a volume of 25.0 ml and a mass of 100 g. what is the density of the object
Tpy6a [65]
Density= mass/volume

             = 100/25
density  = 4g/ml
8 0
3 years ago
What do protons determine about an element
Daniel [21]
The number of protons in an atoms determines the atoms identity. Electrons determine the electrical charge.
5 0
3 years ago
Read 2 more answers
HELP ASAP !!!!!
Anna35 [415]

Answer:

Explanation:

Take a random sample of nuts from the jar.  Let's take two handfuls, after shaking the jar and mixing the nuts thoroughly.  Separate the nuts into almonds and cashews.  Count each pile, then do the following calculation (these numbers are random, for example only).

                <u> Count</u>   <u>Percentage %</u>

Almonds       38        (38)/(87)x100

Cashews      <u> 49</u>         49/87x100

                      87          87/87 = 100%

Ratio of Almonds to Cashews:  <u>38/49</u>

3 0
2 years ago
How can you tell that a gas is a halogen? ___?
OLEGan [10]
If it is located at the second to last row of the periodic table (the halogen family), has seven electrons on it's outer shell, and has an oxidation number of -1, it is a halogen.
Hope this helps : D
8 0
3 years ago
Typical "hard" water contains about 2.0 x 10–3 mol of Ca2+ per liter. Calculate the maximum concentration of fluoride ion that c
malfutka [58]

Answer:

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Explanation:

Hello,

In this case, for the described situation, we infer that calcium reacts with fluoride ions to yield insoluble calcium fluoride as shown below:

Ca^{+2}(aq)+2F^-(aq)\rightleftharpoons CaF_2(s)

Which is typically an equilibrium reaction, since calcium fluoride is able to come back to the ions. In such a way, since the maximum amount is computed via stoichiometry, we can see a 1:2 mole ratio between the ions, therefore, the required maximum amount of fluoride ions in the "hard" water (assuming no other ions) turns out:

[F^-]_{max}=2.0x10^{-3}\frac{molCa^{2+}}{L}*\frac{2molF^-}{1molCa^{2+}}  \\

[F^-]_{max}=4x10{-3}\frac{molF^-}{L}

Best regards.

8 0
3 years ago
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