All categories

2 years ago

Plz help me write on ur own words and ignore where it says pictures

Chemistry

1 answer:

Alinara [238K]2 years ago

7 0

Answer:

· Matter is anything that takes up space.

· A liquid is anything that isn't a solid or gas.

· Gas is anything that isn't a solid or liquid.

· A molecule is when chemically two or more atoms come together.

Explanation:

<h3>Hope this Helps!!</h3>

You might be interested in

How much heat is absorbed in the complete reaction of 3.00 grams of SiO2 with excess carbon in the reaction SiO2(g) + 3C(s) → Si

defon

Answer:

31.24 kJ

Explanation:

SiO₂(g) + 3C(s) → SiC(s) + 2CO(g) ΔH° = 624.7 kJ/mol

First we <u>convert 3.00 grams of SiO₂ to moles</u>, using its <em>molar mass</em>:

3.00 g SiO₂ ÷ 60.08 g/mol = 0.05 mol

Now we <u>calculate the heat absorbed</u>, using the <em>given ΔH°</em>:

If the complete reaction of 1 mol of SiO₂ absorbs 624.7 kJ, then with 0.05 mol:

0.05 mol * 624.7 kJ/mol = 31.24 kJ of heat would be absorbed.

6 0

2 years ago

Energy levels of the Atom O

Artyom0805 [142]

Answer:

Oxygen has 2 energy levels

Explanation:

4 0

2 years ago

The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P

Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

Pb: 207.2;
S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$ . There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$ :

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$ .

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$ .

There's one mole of $\rm Pb$ in each mole of $\rm PbS$ . There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$ ?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$ .

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$ .

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$ ?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$ .

5 0

3 years ago

PLEASE HELP ME

vichka [17]

Light moves in a straight line except at surfaces between different transparent materials, where its path bends.

4 0

3 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

3 years ago