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meriva
3 years ago
10

What is the name for Si3S

Chemistry
1 answer:
sesenic [268]3 years ago
7 0

Answer:

Silicopropane

Explanation:

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put the contributions to the understanding of the atomic structure in order from most recent at the top to the earliest at the b
r-ruslan [8.4K]

Answer:

From Top to Bottom:

- Democritus coming up with the concept of an atom

- Dalton discovering that atoms are the smallest part of an element

- Rutherford discovering the nucleus of an atom

- Thomson discovering electrons

- Bohr modeling electrons orbiting the nucleus

- Schrodinger modeling electrons in the electron cloud

Explanation:

The best way to think about this is from the inside out. Democrats (who lived long before any of the other scientists mentioned) was the one who thought of the idea of the atom. - Therefore, this must be first because all other choices are elaborations on the idea that atoms exist. Next must be Dalton. Dalton saw atoms as "cannonballs" if you will; a solid mass. So then after that, Rutherford and his gold foil experiment (he discovered that some rays he shot through gold foil were deflected back; ie the existence of concentrated areas in an atom, ie the nucleus). Then we get into the information on electrons. We must start with discovery (Thomson). Heres where it gets complicated. Electrons don't <em>actually </em>orbit the nucleus, they exist in electron clouds. So it would be Bohr, who came up with the idea that electron exist outside the nucleus, then Schrodinger, who elaborated on Bohr's theory. Hope this helps!

Nat, Junior

Accel + AP Chem student

5 0
1 year ago
What is the treatment and disposal of Mercury<br>​
PIT_PIT [208]
<h2>Hey there! :) </h2>

<h3>The treatment and disposal of Mercury:</h3>

  • Heating and incineration can release the mercury vapor into atmosphere causing atmospheric pollution. The process of solidification and disposal into secured landfill, gas phase recovery of mercury, and thermal treatment is gaining interest in mercury treatment and recovery field by various researchers and industries.
<h2>HOPE IT HELP YOU </h2>
4 0
2 years ago
Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low
LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

4 0
3 years ago
Name three of the most sensitive parts of your body (Appropriate parts. Ex: finger, wrist, shoulder, etc.)
yKpoI14uk [10]
Inner thigh, eyes, brain
3 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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