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shutvik [7]
3 years ago
15

When you pull on a window shade,you exert what?

Chemistry
1 answer:
yulyashka [42]3 years ago
6 0
Force... I hope this helped you
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Give the oxidation state of sulfur in (d) H₂S ;
earnstyle [38]

The correct answer is -2 oxidation state.

Assume that the oxidation number of the sulfur atom in H2S S is x.Now, 2+x=0 ⇒ x=−2. As a result, the oxidation number of sulfur in H2S is -2. Various organosulfur compounds are produced by combining Hydrogen sulfide with sulfur. Sewer gas, swamp gas, stink damp, and sour damp all include Hydrogen sulphide, an invisible gas. It smells strongly of eggs, even in small amounts. It is extremely flammable and poisonous. While drilling for and producing crude oil and natural gas, gases like Hydrogen sulfide (H2S) are regularly found.

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5 0
2 years ago
Manual handling of materials accounts for the primary source of injury in the workplace
irinina [24]
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7 0
3 years ago
Which subatomic particle of an atom never changes?
AlexFokin [52]
Protons are one of the fully stable parts of matter
8 0
4 years ago
The polyatomic nitrate ion (NO3−) is present in both the products and the reactants of the chemical equation shown below.
Diano4ka-milaya [45]

Answer:

There is one nitrate ion in the products, and two in the reactants of this equation.

Explanation:

1: We know a nitrate ion has the formula NO_{3}^{- , so we just need to count how many of them are on each side of the equation.

2: To find how many are in the reactants of the equation, you look at the left hand side (before the arrow). You can see the section (NO_{3} )_{2} , which shows that there are two nitrate ions in the reactant side (as seen by the little 2).

3: To find how many are on the products side, you do the same thing - this time there is only one NO_{3}^{- in the lithium nitrate.

So, there is one nitrate ion in the products, and two in the reactants of this equation.

7 0
3 years ago
Read 2 more answers
A sample of gas with a mass of 21.3 g is confined to a vessel of volume 7.73 L at 0.880 atm and 30.00C.
rodikova [14]

Answer:

(a) 77.9 g/mol

(b) 3.18 g / L

Explanation:

<u>(a)</u> We need to use the ideal gas law, which states: PV = nRT, where P is the pressure, V is the volume, n is the moles, R is the gas constant, and T is the temperature in Kelvins.

Notice that we don't have moles; we instead have the mass. Remember, though that moles can be written as m/M, where m is the mass and M is the molar mass. So, we can replace n in the equation with m/M, or 21.3/M. The components we now have are:

- P: 0.880 atm

- V: 7.73 Litres

- n: m/M = 21.3 g / M

- R: 0.08206

- T: 30.00°C + 273 = 303 K

Plug these in:

PV = nRT

(0.880)(7.73) = (21.3/M)(0.08206)(303)

Solve for M:

M = 77.9 g/mol

<u>(b)</u> The equation for the molar mass is actually:

M = (dRT)/P, where d is the density

We have all the components except d, so plug them in:

77.9 = (d * 0.08206 * 298) / 1

Solve for d:

d = 3.18 g / L

3 0
3 years ago
Read 2 more answers
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