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3 years ago

Changes in a state of matter include?

Chemistry

2 answers:

o-na [289]3 years ago

4 0

Answer: melting, freezing, sublimation, deposition, condensation, and vaporization

Pepsi [2]3 years ago

4 0

Answer:

SOLID,LIQUID.GAS.,PLASMA

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When a balloon containing 635 mL of air is taken from sea level (at standard pressure) to a higher altitude, the balloon expands

Fiesta28 [93]

Answer:

0.766atm

Explanation:

Boyles law

P1V1=P2V2

1×635=P2×829

P2=635/829

P2=0.766atm

8 0

3 years ago

Redi placed pieces of meat in several jars. He divided the jars into two groups. He covered the first group of jars with fine cl

OlgaM077 [116]

Answer:

Experiment Question, Independent & Dependent Variable, Null & Alternate Hypothesis

Explanation:

Experiment Question : Whether fine cloth coverage over chicken jars effect flies & their eggs in ambience ?

Independent Variable : The causal variable, ie cloth cover existence on the chicken jar

Dependent Variable : The resultant variable, ie presence of flies & their eggs near jar, meat

Null Hypothesis [H0] : Cloth cover on chicken jar doesn't effects flies' & eggs' presence around jar, chicken.

Alternate Hypothesis [H1] : Cloth cover on chicken jar effects flies' & eggs' presence around jar, chicken.

6 0

3 years ago

0.0500 mol of gas occupies a cylinder which is sealed on top by a moveable piston. The piston is circular, with a mass of 30.0 k

Viefleur [7K]

Answer:

The workdone by both N₂ and neon gas is 49.3 J

The change in internal energy of N₂ and neon gas is 125.6 J and 73.54 J respectively

The heat for N₂ and neon gas is 171.9 J and 122.84 J respectively.

Explanation:

Given that:

number of moles = 0.05 mole

mass of the piston = 30 kg

diameter = 5.00 cm = 0.05 m

Area (A) = πr²

$Area (A) = \pi*(\frac{0.05}{2})^2 \\ \\ Area (A) = 0.0019635 \\ \\ Area (A) = 19.635*10^{-4} \ m^2$

The piston is said to move from 30 cm - 40 cm

So, the change in volume ΔV is calculated as:

$=(40-30)*10^{-2} *19.635*10^{-4}$

$= 1.9635*10^{-4} \ m^3$

Outside the cylinder; the pressure $P_{air}= 1 \ atm$ = 101325 Pa

Thus, workdone $w_1$ = PΔV

= $101325*1.9635*10^{-4}$

= 19.90 J

The gravitational work $w_2 = mgh$

Given that the height (h) = 10 cm = 0.1 m

Then; $w_2 = 30*9.8*0.1$

$w_2 = 29.4 \ J$

The total workdone $w_{total}$ for both cases is:

$w_{total } =w_1 + w_2$

$w = (19.90 + 29.4) \ J$

$w =49.3 \ J$

The pressure of gas inside the cylinder is determined as:

$P_{in}.A = P_{out}.A +mg$

$(P_{in}-P_{out}) = \frac{mg}{A} \\ \\ P_{in} -10^5 = \frac{30*9.8}{19.635*10^{-4}} \\ \\ P_{in} = 149732.6203+10^5 \\ \\ P_{in} = 2.497*10^5 \ Pa$

a). assuming that the gas is N₂.

$C_v =\frac{5}{2}R$

Thus, the change in internal energy ΔU is given as:

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{5}{2}R \delta T \\ \\ \delta U = \frac{5}{2}nR \delta T$

Since $P_{in} \delta V = nR \delta T ; \ Then$ ;

$\delta \ U = \frac{5}{2} P_{in} \delta V \\ \\ \delta \ U = \frac{5}{2}*2.497*10^5 *1.9635*10^{-4} \\ \\ \delta \ U = 122.57 \ J$

ΔU ≅ 125.6 J

The heat Q = ΔU + W

Q = (122.6 + 49.3) J

Q = 171.9 J

b) In Neon gas:

$C_v = \frac{3}{2}R$

∴

change in internal energy is;

$\delta U = nC_v \delta T \\ \\ \delta U = n* \frac{3}{2}R \delta T \\ \\ \delta U = \frac{3}{2}P_{in}.V$

$\delta U = \frac{3}{2}*2.497*10^5*1.9635*10^{-4}$

ΔU = 73.54 J

The heat Q = ΔU + W

Q = (73.54 + 49.3) J

Q = 122.84 J

5 0

3 years ago

Does mgso4 dissolve via ionization

3241004551 [841]

I dont think so......

4 0

4 years ago

Write the name of this structure

faltersainse [42]

Answer:

The name of the compound is:

4–bromo–5,5–dichloro–hex–3–ene–1–yne

Explanation:

To the above compound, we simply do the following:

1. Identify the functional group of the compound.

2. Locate the longest continuous carbon chain. This gives the parent name of the compound.

3. Identify the substituent group attached.

4. Give the substituents the lowest possible count alphabetically.

5. Combine the above to obtain the name of the compound.

Now, we shall determine the name of the compound as follow:

1. The compound has carbon to carbon triple bond (C≡C) yne and carbon to carbon double bond (C=C) ene.

2. The longest continuous carbon chain is 6 i.e hexane.

3. The substituents group attached are:

i. Bromine (bromo), Br

ii. Chlorine (Chloro), Cl

4. In counting, ene comes before yne. But in this case, ene assumes same position from either side, So we shall consider counting from the side that gives yne the lowest count. Thus, ene is at carbon 3 and yne is at carbon 1. Therefore,

i. Bromo, Br is at carbon 4.

ii. Chloro, Cl appears twice and both are at carbon 5.

5. The name of the compound is:

4–bromo–5,5–dichloro–hex–3–ene–1–yne.

7 0

4 years ago