how do you separate a heterogenous mixture that contains powdered sulfur, pebbles, powdered iron, sand, and magnesium sulfate??

The image shows the representation of an unknown element in the periodic table. A square is shown. Inside the square twelve is w

avanturin [10]

Answer:

The combined mass of all the protons and electrons is 24.305.

Explanation:

From the information given about the atom, we can see that the relative atom mass of the element is 24.305.

The relative atomic mass of an atom is the combined mass of all its isotope in proportion of their geonormal abundances. On the periodic table, this mass deals with the amount of protons and neutrons present in a given atom.

7 0

3 years ago

The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the rea

timofeeve [1]

Answer:

$K_{p}$ for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure ( $K_{p}$ ) for this reaction can be written as-

$K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}$

where $P_{NO_{2}}$ and $P_{N_{2}O_{4}}$ are equilibrium partial pressure of $NO_{2}$ and $N_{2}O_{4}$ respectively

Hence $K_{p}=\frac{(0.095)^{2}}{(0.0005)}$ = 18.05

So, $K_{p}$ for the reaction is 18.05

3 0

3 years ago

How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)

user100 [1]

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

4 0

3 years ago

Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

Best regards.

4 0

3 years ago

What is the solution to the problem expressed to the correct number of significant figures

quester [9]

I don't know what the problem is, but here are some rues to help you out:

All non-zero figures are significant
When a zero falls between non-zero digits, that zero is significant.
When a zero falls after a decimal point, that zero is significant.
When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
When adding and subtracting, the answer is limited to the number of decimal places in the number with the least number of decimal places.

4 0

3 years ago