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Elenna [48]
3 years ago
5

Has the spring been stretched beyond its elastic limit?Explain your answer​

Chemistry
1 answer:
lara31 [8.8K]3 years ago
4 0
Hahajjajajajajajhsjajajaiaiajauuaa
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Prepare a 1.00 L of M solution of sodium carbonate . Determine the amount of sodium carbonate needed incorrect significant figur
Vedmedyk [2.9K]

Answer:b

Explanation:

4 0
3 years ago
The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o
choli [55]

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

4 0
3 years ago
Read 2 more answers
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
What is the mass of a liquid having a density of 1.40 g/mL and a volume of 3.090 L
riadik2000 [5.3K]
It has a mass of 4.326 grams. 


4 0
3 years ago
Read 2 more answers
2. Some nitrogen at a pressure of 35.75 p.s.i is in a 100 L container. If the container's volume is reduced to 2250 ml then what
elena-s [515]

Answer:

1455.6

Explanation: you first convert 2250ml to l by dividing by 1000 so you get 2.25l then you use Boyles law which is p1v1=p2v2 then insert values

35.75*100=p2*2.25 then divide both sides by 2.25 then you get 1455.6

4 0
3 years ago
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