All categories

3 years ago

Please help! BRAINLIEST to right answer!!!

Chemistry

1 answer:

Shalnov [3]3 years ago

5 0

Answer:

Explanation:

You might be interested in

Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper

xxTIMURxx [149]

Answer:

Explanation:Use the Gizmo to mix 200 g of copper at 100 °C with 1,000 g of water at 20 °C. Record the data and calculated answers for copper in the 2 tables below. Accepted values for % error calculations can be found below these 2 tables.

DATA

Copper

Lead

Mass of Metal

3 0

3 years ago

How many joules would be absorbed for the water in Problem 1

9966 [12]

Answer:

5000 cal X 4.184 J/cal =

Explanation:

20,920 jules

8 0

4 years ago

Convert to moles: show work

Ket [755]

I’m just letting you know this is really easy you just calculate the molar mass of each compound and divide the amount of the compound (grams) by the molecular Mass

6 0

3 years ago

What are carbon atoms made of?

juin [17]

Atoms with six protons and six neutrons — carbon. Carbon is a pattern maker. It can link to itself, forming long, resilient chains called polymers. It can also bond with up to four other atoms because of its electron arrangement.

3 0

3 years ago

Read 2 more answers

A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the m

aliya0001 [1]

Answer:

311.25k

Explanation:

The question assumes heat is not lost to the surroundings, therefore

heat emitted from hotter sample ( $q_{\ lost}$ )= heat absorbed by the less hotter sample( $q_{\ gain}$ )

The relationship between heat (q), mass (m) and temperature (t) is $q = mc\Delta t$

where c is specific heat capacity, $\Delta t$ temperature change.

$\Delta t$ = $t_{\ final} - t_{\ initial}$

equating both heat emitted and absorb

$-q_{\ lost} = q_{\ gain}$

$-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})$

where the values with subset 1 are the values of the hotter sample of water and the values with subset 2 are the values of the less hot sample of water.

C will cancel out since both are water and they have the same specific heat capacity.

so we have

$-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})$

where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?

-50 * ( $t_{final}$ - 330) = 30 * ( $t_{final}$ - 280)

-50 $t_{final}$ + 16500 = 30 $t_{final}$ - 8400

80 $t_{final}$ = 16500+8400

80 $t_{final}$ = 24900

$t_{final}$ = 24900/80 = 311.25k

8 0

3 years ago