Which of the following actions does not require power

A depression that receives more water then is we will exist as a stream for a long period of time

Anvisha [2.4K]

The answer is the lake.

A lake is a large expanse of terrestrial water, consisting of streams that are linked to the lake. Although some dimensions of some salt-water lakes are considered island seas; they are therefore distinct from the lagoons, and are larger and deeper than the ponds, while remaining a water body by definition.

7 0

3 years ago

How many liters of solution will contain 1.5 moles of CaCl2 if the solution is 6.0 M CaCl2?

sashaice [31]

Answer:

0.25

Explanation:

volume= number of moles over concentration

hence v=1.5/6

6 0

3 years ago

Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o

Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M$

where,

$n_p$ = number of protons

$m_p$ = mass of one proton

$n_n$ = number of neutrons

$m_n$ = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is $_{15}^{31}\textrm{P}$

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

$\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399$

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0

2 years ago

An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density

V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

$Density = \dfrac{4 \times M_w}{N_A \times a^3}$

where

$M_w$ = molar mass of the compound

$N_A=$ avogadro's constant

$a^3 =$ the volume of a unit cell

Given that:

Density $(\rho)$ = 19.30 g/cm³

a = 0.408 nm

a = $0.408 \times 10^{-9} \times 10^{2} \ cm$

a = $4.08 \times 10^ {-8} \ cm$

∴

$19.3 = \dfrac{4 \times M_w}{(6.023 \tmes 10^{23})\times (4.08 \times 10^{-8})^3}$

$M_w= \dfrac{19.3\times (6.023 \times 10^{23})\times (4.08 \times 10^{-8})^3}{4}$

$M_w=197.37 \ g/mol$

Thus, the molar mass of 197.37 g/mol element is Gold (Au).

4 0

3 years ago

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

muminat

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ?

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

$n = \dfrac{m}{MM}$

$5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}$

$m = 5.20*84.2$

$\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark$

_______________________

I Hope this helps, greetings ... Dexteright02! =)

4 0

3 years ago

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