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Darina [25.2K]
3 years ago
11

Calculate the sum of 6.078 g and 0.3329 g

Chemistry
1 answer:
Gelneren [198K]3 years ago
3 0
The sum means that you need to add the two numbers together. Therefore, the answer (or sum) should be 6.4109.
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What are the electrons for calcium
IRINA_888 [86]
The traditional calcium atom has twenty protons and twenty electrons making it neutral.


The calcium in the pic is a calcium ion so the number of protons and electrons are not equivalent.


Since it's 2+ that means the ion is positively charged and for that to happen electrons are away.

So 20-2=18

There are 18 electrons
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Is volcanoes eruption rapid change or slow change​
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Fast change

Explanation:

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8 0
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8 0
3 years ago
A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First
bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0
3 years ago
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