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sergejj [24]
3 years ago
9

An object submerged into 50 cm^3 of water raises the water level to 65 cm^3. The mass of the object is 30 g. What is its density

?
Chemistry
1 answer:
Anna007 [38]3 years ago
8 0

Answer:

Convert the volume to m³.

Explanation:

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How many gram-atoms are in 120-gram sample of calcium metal? How many atoms is this?
Elden [556K]
(sample g/1) X (1 mole/40.078(MW of Ca)) = moles of sample (moles of sample)(6.022 x 10^23( no of atoms)/ 1 mole) = # of atoms in a 120 g sample of calcium Avogadro's number=6.022x 10^23 atoms in 1 mole
3 0
3 years ago
If you had 15 molecules of H2 and an unlimited supply of N2, how many
Masja [62]

Answer:

10 molecules of NH₃.

Explanation:

N₂ + 3H₂ --> 2NH₃

As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:

  • 15 molecules H₂ * \frac{2moleculesNH_3}{3moleculesH_2} = 10 molecules NH₃

10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.  

8 0
3 years ago
What is the frequency of a photon that has a wavelength of 754 um?​
Bogdan [553]

Answer:

0.39760273

Explanation:

I typed into calculator hope it's right.

6 0
3 years ago
Which of the following is not a characteristic of a covalent compound?
TEA [102]
<span>'It is formed when metal atoms lose electrons to nonmetal atoms' is the incorrect statement. This statement is the definition of ionic bonding.

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3 0
3 years ago
How much heat is absorbed when a 298.3 g piece of brass goes from 30.0 to 150
igor_vitrenko [27]

Answer:

Q = 1360.248 j

Explanation:

Given data:

Mass of brass = 298.3 g

Initial temperature = 30.0°C

Final temperature = 150°C

Specific heat capacity of brass = 0.038 J/g.°C

Heat absorbed = ?

SOLUTION:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30.0°C  

ΔT = 120°C

Q = 298.3 g × 0.038 J/g.°C × 120°C

Q = 1360.248 j

3 0
3 years ago
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