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jeyben [28]
3 years ago
11

An isotope of iron has 28 neutrons. If the atomic mass of the isotope is 54, how many protons does it have?

Chemistry
2 answers:
Crazy boy [7]3 years ago
5 0
26 because you need to do 54-28=26
The atomic mass number is the number of protons AND neutrons together.
elena-14-01-66 [18.8K]3 years ago
4 0
An isotope is a varying in the neutrons because the protons NEVERRRR change!!

54 - 28 = <span>26 
</span>
Answer: 26


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Here ya go if you need the link we’re I found this answer key https://studylib.net/doc/8708211/exam-2---chemistry

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3 years ago
What is the pH if the [H+] concentration is 3 x10^-13​
Shkiper50 [21]

Answer:

pH = 12.52

Explanation:

Given that,

The [H+] concentration is 3\times 10^{-13}.

We need to find its pH.

We know that, the definition of pH is as follows :

pH=-log[H^+]

Put all the values,

pH=-log[3\times 10^{-13}]\\\\pH=12.52

So, the pH is 12.52.

6 0
3 years ago
If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
QveST [7]
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
4 0
3 years ago
Heat transfer putting a heating pad on leg
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8 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
3 years ago
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