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lapo4ka [179]
3 years ago
10

Which sentences in the passage are true?

Chemistry
1 answer:
user100 [1]3 years ago
7 0

Answer:

First and last sentence .‍♀️

Explanation:

An important thing to keep in mind about the Reading Comprehension section of the GRE as we use PowerPrep online to study is that it is just that—reading comprehension. In other words, as difficult as it may seem, and it can be pretty tricky, the test makers will always give us all the information we need in the passage to answer the question.

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3. Determine the pH of each of the following solutions. (Hint: See Sample Problem B.) a. 1.0 x 10-2 M HCI c. 1.0 x 10-MHI b. 1.0
densk [106]

Answer:

a. pH = 2 b. pH = 3 c. pH = 1 d. Unanswerable

Explanation:

pH = -log[H+] OR pH = -log{H3O+]

                    and inversely

pOH = -log[OH-]

1. Determine what substance you are working with, (acid/base)

2. Determine whether or not that acid or base is strong or weak.

a. 1.0 x 10^-2M HCl

HCl is a strong acid, therefore it will dissociate completely into H+ and Cl- with all ions going to the H+, therefore, the concentration of HCl and concentration of H+ are going to be equal, meaning we simply take the negative logarithm of the concentration of HCl and that would equal pH

pH = -log[H+]

pH = -log(1.0x10^-2)

pH = 2

b. 1.0 x 10^-3M HNO3

HNO3 like part a, is a strong acid, therefore it would simply require you to take the negative logarithm of the concentration of the compound itself, to find its pH.

pH = -log[H+]

pH = -log(1.0 x 10^-3)

pH = 3

c. 1.0 x 10^-1M HI

Like the previous parts, HI is a strong acid

pH = -log[H+]

pH = -log(0.10)

pH = 1

d. HB isn't an element, nor is it a compound so that would be unanswerable.

3 0
3 years ago
Please I need this help me​
BaLLatris [955]

Answer:

first of all is that really chemistry

3 0
3 years ago
Calculate the radius of a tantalum (ta) atom, given that ta has a bcc crystal structure, a d ensity of 16.6 g/cm3, and an atomic
Sidana [21]

solution:

Ta is BCC, a=\frac{4}{\sqrt{3}r},n=2atoms/cell\\p=16.6gram/cm^3,A=180.9gram/mol\\P=\frac{A\times n}{N_{a}\times v}=\frac{180.9\frac{gram}{mol}\times 2 \frac{atom}{cell}}{6.023\times10^23\frac{atom}{mol}\times(\frac{4}{\sqrt{3}\times r})^3\frac{cm^3}{cell}}\\=16.6\frac{gram}{cm^3}\\r=(\frac{180.9\times2}{16.6\times6.023\times10^23\times(\frac{4}{\sqrt{3}})^3})^\frac{1}{3}\times10^8


4 0
4 years ago
The atom in a sample of an element must contain nuclei with the same number of?
slava [35]
Hey there! 

Answer: Protons 

The atom in a sample of an element must contain nuclei with the same number of protons. If the protons change in number, then the element will modify as well. 

Thank you!
7 0
3 years ago
A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the
alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.4 atm

P_2 = final pressure of gas = 1.0 atm

V_1 = initial volume of gas = 250 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 15^oC=273+15=288K

T_2 = final temperature of gas = 27^oC=273+27=300K

Now put all the given values in the above equation, we get:

\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}

V_2=625mL

Therefore, the volume of the bubble is, 625 mL

7 0
3 years ago
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