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Vlad [161]
3 years ago
9

What type of bonding holds the atoms in an h2o molecule together?

Chemistry
1 answer:
Ivanshal [37]3 years ago
3 0
Covalent bond  two non metals
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Va rog ! urgent! E pt chimie ​
n200080 [17]

Answer:

An element

It stays shiny

Explanation:

Pure Gold is an element.

An element is a distinct substance that cannot be split-up into simpler substances. Such substances consists of only one kind of atom.

There are over a hundred elements known to date.

As an element, Gold is classified as a metal due to its very unique set of properties.

One of the indicator that gold does not react with oxygen is that it stays shiny. It does not give rusty look when exposed to air.

Substances that combines with oxygen have a rusty look or change appearance when expose to air. For example, iron.

5 0
3 years ago
90 grams of NaNO3 were added to solution at 10 degrees Celsius. What type of solution was formed?
vichka [17]

Answer:

friend me on here and imma send you the link Explanation:

4 0
3 years ago
Please help <br><br> 10 POINTS!!!
salantis [7]
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5 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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