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3 years ago

What is the mass number of 80Br

Chemistry

1 answer:

Amanda [17]3 years ago

4 0

Mass number of an element write in bottom of it!

So, there 80 would be the mass number of ₈₀Br.

Hope this helps!

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For the following molecules, what is the order of Bronsted acidity, from highest to lowest

tankabanditka [31]

Answer:

Rank these compounds in order of increasing basicity. (Least basic to most basic)

a. C, B, A

b. B, A, C

c. C, A, B

d. A, B, C

ANSWER: d

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Explanation:

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4 0

2 years ago

If you are given only the mass number of neutrons, how do we find an atomic number?

77julia77 [94]

Mass number is the sum of protons and neutrons.

Atomic number is the total number of protons present.

Mass number = protons + neutrons

Atomic number = protons present.

From the above equations we can conclude that,

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Thus we can find out the atomic number by subtracting the number of neutrons from the mass number.

3 0

2 years ago

Read 2 more answers

Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil

quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

$Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}$

7 0

2 years ago

Gt4rgbrbrtbr\t?<br>frehjkgbrtnvboror

3241004551 [841]

Answer:I can't see any question

8 0

2 years ago

Im hvaing a hard time getting the right answer

Tanya [424]

Answer:

$V=23.9mL$

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

$n=2.49g*\frac{1mol}{125.55 g}=0.0198mol$

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

$M=\frac{n}{V}\\\\V=\frac{n}{M}$

By plugging in the moles and molarity, we obtain:

$V=\frac{0.0198mol}{0.830mol/L}=0.0239L$

Which in mL is:

$V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL$

Best regards!

6 0

2 years ago