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Marta_Voda [28]

2 years ago

9

An inflated balloon is left outside overnight. initially it has a volume of 1.84 L when the temperature is 293.4 K and the press

ure. what temperature will the balloon have a volume of 1540mL. If the pressure falls to 14.41 psi?

1 answer:

Mrac [35]2 years ago

5 0

Answer:

ffff

Explanation:

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Addie mixes together a solution of 10 grams of

Gnesinka [82]

Answer: 150 grams

Explanation: m = V × ρ

= 15 milliliter × 10 gram/cubic centimeter

= 15 cubic centimeter × 10 gram/cubic centimeter

= 150 gram

5 0

3 years ago

Would an opah fish be better suited to live in cold water or warm water? explain why

Kaylis [27]

Don't blame me if you get it wrong i'm a dumb but yet successful student

Warm blood gives deep-sea fish a boost, according to Wegner. The opah's muscles and nervous system likely function faster than an equivalent fish with cold blood. ... This fish, the southern opah, lives in colder waters than the northern opah, so it would be harder to keep warm, Wegner said — but even more beneficial.

4 0

3 years ago

What are the two types of transmutations?

joja [24]

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7 0

2 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

3 years ago

Draw a lewis structure for ketene, c2h2o, which has a carbon–carbon double bond

MrMuchimi

Visual representation of covalent bonding indicating the valence shell electrons in the molecule, lines represents the shared pair of electron and pair of electrons that are not involved in bonding are represented as dots(lone pairs) are known as Lewis structures.

Compound formation takes place in order to complete the octet of each element that is according to octet rule, each atom forms bond with other atom in order to complete their octet that is to get eight electrons in its valence shell and attain stability.

An organic compound of the form $R^{'}R^{''}C=C=O$ is known as ketene.

The given ketene is $C_2H_2O$ .

The number of valence electron of:

$C = 4$

$H = 1$

$O = 6$

The number of valence electrons in $C_2H_2O$ = $2\times 4+2\times 1+1\times 6 =16$

2 electrons are involved in each single bond between carbon and hydrogen and 4 electrons are involved in each double bond formed between carbon-carbon and carbon-oxygen. Hence, the total number of electrons involved in bond formation are 12 and rest 2 pair of electrons are present on oxygen as lone pair of electrons.

Therefore, the attached image is the Lewis structure of $C_2H_2O$ .

8 0

3 years ago