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3 years ago

Is any one in connections academy inedd unit 5 unit test answrs now

Chemistry

1 answer:

Usimov [2.4K]3 years ago

7 0

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What are the factors that affect the rate of dissolution

LUCKY_DIMON [66]

<span>The amount of a component or solute that dissolves in a given volume of solvent at a certain temperature. </span>

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3 years ago

A gas is contained in a tick walled balloon. When the pressure changes from 2. 95atm to ___atm, the volume changes from 7. 456L

tia_tia [17]

A gas is contained in a tick-walled balloon. When the pressure changes from 2. 95atm to <u>2.57</u> atm. p.

<h3>What is pressure?</h3>

Pressure is a force perpendicular to the unit area on which it is applied.

Given that, P1, the initial pressure is 2.95 atm

The initial volume, V1 = 7.456 L

The final volume, V2 = 4.782 L

The initial temperature, T1 = 379k

The final temperature, T2 = 212k

The equation will be

$\dfrac{P_1V_1}{T_1 } =\dfrac{P_2V_2}{T_2}$

Putting the values in the equation

$\dfrac{2.95 \times 7.456 L }{379k } =\dfrac{P_2\times 4.782 L}{212k}\\\\\\P_2 = \dfrac{2.95 \times 7.456 L \times 212k }{379k \times 4.782 L} = 2.57\;atm.$

Thus, the pressure changes from 2. 95atm to <u>2.57</u> atm p.

Learn more about pressure

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2 years ago

CoSO₄ and Pb(NO₃)₂ are mixed in aqueous solution.

laila [671]

Answer:

See below.

Explanation:

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3 years ago

A surface wave is a combination of what two waves

Alika [10]

transverse wave and a longitudinal wave

Hope it helps have a blessed day!:)

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3 years ago

Read 2 more answers

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

6 0

3 years ago