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3 years ago

6

The long-term weather patterns that are typical in a location are the location's___?

1 answer:

fenix001 [56]3 years ago

3 0

Climate is an area's long-term weather patterns

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How many elements are in N2 and how many atoms are in N2?

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Answer:

2

Explanation:

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2 years ago

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Isotope b has a half-life of 3 days. a scientist measures out 100 grams of this substance. after 6 days has passed, the scientis

Brilliant_brown [7]

So half life is the time taken for a sample to decay to half its original mass, its a constant and applies to any original mass, it could be 5g or 1kg, it will take the same amount of time for the original mass to half. In this case the half life is 3 days.

After 3 days the sample will be at half its original mass, now 50g.

Now we can treat the 50g as if its a new sample. After another 3 days (6 days in total) there will be half of 50g left, = 25g.

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In an experiment, you use a ruler to measure two plants. What can you do with the data you gather?

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C. is the best choice

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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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3 years ago

What are at least 3 places carbon can be found?

Tcecarenko [31]

Answer:

Carbon can be found in the ocean, rocks, and the atmosphere around us.

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2 years ago