How many significant figure are in the number 0.000850320

a natrual distsater caused a population of 4,695 organisms to migrate to a new habitat. a few generations after the disaster it

Scilla [17]

Answer:

49.54%

Explanation:

Given parameters:

Number of organisms in original habitat = 4695

Number of organism in new habitat = 2326

Solution:

To find the percentage of the organisms in the new habitat that has migrated to the new habitat, we use the expression below:

% of the population in the new habitat = x 100

% of the population in the new habitat = x 100 = 49.54%

7 0

2 years ago

A 0.539 g sample of a metal, M, reacts completely with sulfuric acid according to the reaction M(s)+H2SO4(aq)⟶MSO4(aq)+H2(g) A v

Charra [1.4K]

Answer:

The molar mass of the metal is 54.9 g/mol.

Explanation:

When we work with gases collected over water, the total pressure (atmospheric pressure) is equal to the sum of the vapor pressure of water and the pressure of the gas.

Patm = Pwater + PH₂

PH₂ = Patm - Pwater = 1.0079 bar - 0.03167 bar = 0.9762 bar

The pressure of H₂ is:

$0.9762bar.\frac{1atm}{1.013bar} =0.9637atm$

The absolute temperature is:

K = °C + 273 = 25°C + 273 = 298 K

We can calculate the moles of H₂ using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{0.9637atm \times 0.249L }{(0.08206atm.L/mol.K)\times298K} =9.81 \times 10^{-3} mol$

Let's consider the following balanced equation.

M(s) + H₂SO₄(aq) ⟶ MSO₄(aq) + H₂(g)

The molar ratio of M:H₂ is 1:1. So, 9.81 × 10⁻³ moles of M reacted. The molar mass of the metal is:

$\frac{0.539g}{9.81 \times 10^{-3} mol} =54.9g/mol$

4 0

4 years ago

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of

Alenkinab [10]

Answer:

Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together. After the big bang,<em> the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently</em>.

The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.

6 0

3 years ago

Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that

Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻ E° = 2.38 V oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s) E° = -0.25 V reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s) E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0

3 years ago

Steam initially at 0.3 MPa, 2500 C is cooled at constant volume. (a) At what temperature will steam become saturated vapour? [12

AlladinOne [14]

Answer:

a. 123.9°C

b.

c.

Explanation:

Hello, I'm attaching a picture with the numerical development of this exercise.

a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.

b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.

c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.

** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.

Best regards.

5 0

3 years ago