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2 years ago

When measuring the volume of an object you are finding the?

Chemistry

1 answer:

marta [7]2 years ago

7 0

Answer:

C. Amount of 3 dimensional space the object occupies

Explanation:

Single dimension is a laminar

Amount of 2 dimension is area

Amount of 4 dimension is negligible

$.$

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What color does blue and cyan pigment make?

Fed [463]

Answer:

Teal

Explanation:

Blue lightens the cyan.

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2 years ago

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Question 1 (4 points)

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2 years ago

There are millions of organic compounds but only thousands of inorganic compounds because:______

siniylev [52]

Answer:

There are millions of organic compounds but only thousands of inorganic compounds because:

a. organic compounds were formed by living things.

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c. atoms of elements other than carbon never combine with themselves.

d. carbon atoms can combine with up to four other atoms, including other carbon atoms.

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1 year ago

What type of solid is candle wax

son4ous [18]

Hi Sara

I would say it is an amorphous solid.

Because it's gradually loses shape.

I hope that's help:)

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2 years ago

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"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

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2 years ago