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3 years ago

How did your estimated poputation size compare with the actual population size

Chemistry

2 answers:

musickatia [10]3 years ago

8 0

Answer:

Hope this helps

Explanation:

https://seagrant.whoi.edu/wp-content/uploads/2018/05/ESTIMATING-POPULATION-SIZE-1.pdf

Goshia [24]3 years ago

8 0

Answer:

HE coracle—as I had ample reason to know before I was done with her—was a very safe boat for a person of my height and weight, both buoyant and clever in a seaway; but she was the most cross-grained, lop-sided craft to manage. Do as you pleased, she always made more leeway than anything else, and turning round and round was the manoeuvre she was best at. Even Ben Gunn himself has admitted that she was "queer to handle till you knew her way."

Explanation:

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6. 5.50 x 10- molecules of carbon dioxide to moles.

sergeinik [125]

1.24973017189471 is probably the answer to your equation

4 0

3 years ago

Calcular la presión final de un gas que inicialmente está a 21°C y 0,98 atm. Sabiendo que su temperatura aumenta a 37°C.

KatRina [158]

Answer:

1.03 atm

Explanation:

Primero convertimos 21 °C y 37 °C a K:

21 °C + 273.16 = 294.16 K

37 °C + 273.16 = 310.16 K

Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la ley de Gay-Lussac:

T₁P₂ = T₂P₁

En este caso:

T₁ = 294.16 K

P₂ = ?

T₂ = 310.16 K

P₁ = 0.98 atm

Colocando los datos:

294.16 K * P₂ = 310.16 K * 0.98 atm

Y despejando P₂:

P₂ = 1.03 atm

7 0

3 years ago

Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th

aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = ?

Depression in freezing point = $\Delta T_f$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample = 0.275 mol/kg

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( non electrolyte)

$\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg$

$\Delta T_f=11^oC$

$\Delta T_f=T- T_f$

$T_f=T- \Delta T_f=179^oC-11^oC=168^oC$

168°C is the melting point of your impure sample.

4 0

3 years ago

Proteins are also known as enzymes. True or false

julia-pushkina [17]

Answer:

True

Explanation:

3 0

3 years ago

Read 2 more answers

If 31.6 g of KMnO4 is dissolved in enough water to give 160 mL of solution, what is the molarity?

Zina [86]

Answer:

A. 1.25M

B. 19.98g

Explanation:

A. Data obtained from the question include the following:

Mass of KMnO4 = 31.6 g

Volume = 160 mL

Molarity =..?

We'll begin by calculating the number of mole KMnO4 in the solution. This is can be obtained as follow:

Mass of KMnO4 = 31.6 g

Molar mass of KMnO4 = 39 + 55 + (16x4) = 158g/mol

Number of mole of KMnO4 =..?

Mole = mass /Molar mass

Number of mole of KMnO4 = 31.6/158 = 0.2 mole

Now, we can obtain the molarity of the solution as follow:

Volume = 160 mL = 160/1000 = 0.16L

Mole of KMnO4 = 0.2 mole

Molarity = mole /Volume

Molarity = 0.2/0.16 = 1.25M

B. Data obtained from the question include the following:

Volume = 300mL

Molarity = 0.74 M

Mass of H2C2O4 =..?

First, we shall determine the number of mole H2C2O4. This is illustrated below:

Volume = 300mL = 300/1000 = 0.3L

Molarity = 0.74 M

Mole of H2C2O4 =?

Mole = Molarity x Volume

Mole of H2C2O4 = 0.74 x 0.3

Mole of H2C2O4 = 0.222 mole

Now, we can easily find the mass of H2C2O4 by converting 0.222 mole to grams as shown below:

Number of mole of H2C2O4 = 0.222 mole

Molar mass of H2C2O4 = (2x1) + (12x2) + (16x4) = 2 + 24 + 64 = 90g/mol

Mass of H2C2O4 =..?

Mass = mole x molar mass

Mass of H2C2O4 = 0.222 x 90

Mass of H2C2O4 = 19.98g

5 0

3 years ago