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4 years ago

What physical barrier separated groups of snapping shrimp approximately 3 million years ago?

1 answer:

8 0

The isthmus and the Panama Canal were believed to be the physical barrier that separated groups of snapping shrimps 3 million years ago. In addition, one family of snapping shrimps that got separated is the Alpheidae which produces loud sound when it snaps its pincers.

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What is the mass of 5 miles in ba3n2

Rasek [7]

Answer:

2200 g

Explanation:

Data Given:

no. of moles of Ba₃N₂ = 5 moles

mass of Ba₃N₂ = ?

Solution:

Formula used

no. of moles = mass in grams / molar mass

To find mass rearrange the above equation:

mass in grams = no. of moles x molar mass. . . . . . (1)

molar mass of Ba₃N₂

molar mass of Ba₃N₂ = 3(137.3) + 2(14)

molar mass of Ba₃N₂ = 412 + 28

molar mass of Ba₃N₂ = 440 g/mol

Put values in equation 1

mass in grams = 5 moles x 440 g/mol

mass in grams = 2200 g

So,

mass of Ba₃N₂ = 2200 g

8 0

4 years ago

A voltaic cell is constructed from an Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s) half-cell. The initial concentration of Ni2+

SCORPION-xisa [38]

Explanation:

For what I can see, is missing the concentration of [Ag+] in the half-cell. To calculate it:

Niquel half-cell

Oxidation reaction: $Ni \longrightarrow Ni^{2+}+2 e^-$

$E=E^0 - \frac{R*T}{n*F}*ln(1/[Ni^{2+}])$

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

$E=-0.23V - \frac{8.314*298}{2*96500}*ln(1/0.014M)$

$E=-0.285V$

Silver half-cell

Reduction reaction: $Ag^+ + e^- \longrightarrow Ag$

$E=E^0 - \frac{R*T}{n*F}*ln(1/[Ag+])$

$E_{cell}=E_{red} - E_{ox}$

$E_{red}=1.12 V + (-0.855V)=0.835V$

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

$0.835V=0.8V - \frac{8.314*298}{1*96500}*ln(1/[Ag+])$

$[Ag+]=0.26 M$

3 0

3 years ago

According to the graph, how many mice will be born in week 5 if the trend continues?

dlinn [17]

Okay so the answer to this one is very simple 91

7 0

3 years ago

The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) and Kp=0.174 at 243°C. Calculate the value of Kc for

Mashcka [7]

Answer:

The Kc of this reaction is 311.97

Explanation:

Step 1: Data given

Kp = 0.174

Temperature = 243 °C

Step 2: The balanced equation

N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 3: Calculate Kc

Kp = Kc *(RT)^Δn

⇒ with Kp = 0.174

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 Latm/Kmol

⇒ with T = the temperature = 243 °C = 516 K

⇒ with Δn = number of moles products - moles reactants 2 – (1 + 3) = -2

0.174 = Kc (0.08206*516)^-2

Kc = 311.97

The Kc of this reaction is 311.97

3 0

3 years ago

What mass of a 60.0-gram sample of N-16 will remain unchanged after 28.8 seconds?

Yanka [14]

The answer is 3.75 g.

Half-life is the time required for the amount of a sample to half its value.
To calculate the fraction of the sample remained unchanged, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

The half-life of N-16 is 7.13.
So, we know:
t = 28.8 s
 $t_{1/2}$ = 7.13 s

We need:
n = ?
x = ?

We could first use the second equation, to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{28.8 s}{7.13 s}$
⇒ $n=4.04$
⇒ n ≈ 4

Now we can use the first equation to calculate the remained amount of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^4$
⇒ $x= \frac{1}{16}$

If the fraction of the sample is 1/16 = 6.25%, then the mass of the sample could be calculated as:
x g : 6.25% = 60 g : 100%
x =

7 0

3 years ago