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Ray Of Light [21]
3 years ago
5

Please help me, Thank you!

Chemistry
2 answers:
cluponka [151]3 years ago
8 0
Amount of charge your welcome i think
babymother [125]3 years ago
4 0

Answer:

amount of charge

Explanation:

Oxygen and sulfur are both in Group 16, which means they have a -2 charge. They have two more electrons than protons, making the charge of the ion negative.

Hope that helps.

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Which of the following is evidence of a physical change?haora
Arisa [49]
Chopping wood
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molding clay
3 0
3 years ago
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Which of these plants make the enviroment pleasant<br><br> a.sandalwood b.basil c.brahmi
liberstina [14]

Sandalwood

Explanation:

this is the answer please mark me as brainleist

4 0
3 years ago
a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phas
jarptica [38.1K]

Answer:

The equilibrium partial pressure of O2 is 0.545 atm

Explanation:

Step 1: Data given

Partial pressure of SO2 = 0.409 atm

Partial pressure of O2 = 0.601 atm

At equilibrium, the partial pressure of SO2 was 0.297 atm.

Step 2: The balanced equation

2SO2 + O2 ⇆ 2SO3

Step 3: The initial pressure

pSO2 = 0.409 atm

pO2 = 0.601 atm

pSO3 = 0 atm

Step 4: Calculate the pressure at the equilibrium

pSO2 = 0.409 - 2X atm

pO2 = 0.601 - X atm

pSO3 = 2X

pSO2 = 0.409 - 2X atm = 0.297

 X = 0.056 atm

pO2 = 0.601 - 0.056 = 0.545 atm

pSO3 = 2*0.056 = 0.112 atm

Step 5: Calculate Kp

Kp = (pSO3)²/((pO2)*(pSO2)²)

Kp = (0.112²) / (0.545 * 0.297²)

Kp = 0.261

The equilibrium partial pressure of O2 is 0.545 atm

3 0
3 years ago
A boy swings a rubber ball attached to a string over his head in a horizontal, circular path. The piece of string is 1.15 m long
gregori [183]

Answer:

v = 16.49 m/s

Explanation:

Given that,

Length of the string, l = 1.15 m

The ball makes 137 complete turns each minute.

We know that, 1 turn = 6.28 rad

137 turns = 860.79 rad

1 min = 60 s

\omega=\dfrac{860.79\ rad}{60\ s}\\\\=14.34\ rad/s

We need to find the tangential velocity of the ball. It can be given by

v=r\omega\\\\=1.15\times 14.34\\\\v=16.49\ m/s

So, the tangential velocity of the ball is 16.49 m/s.

5 0
3 years ago
2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
5 0
3 years ago
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