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Sveta_85 [38]
4 years ago
9

CaCl2 + Mg3N2 --> how do i balance this?

Chemistry
1 answer:
klemol [59]4 years ago
6 0
Ca3N2+3MgCl2 (double replacement)
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He image shows the representation of an unknown element in the periodic table.
alexandr402 [8]

Answer:

the number of neutrons in an atom of the element is 10.

Explanation:

neutron: mass # -protons

neutron: 19 (18.9984 (rounded)) - 9

neutron: 19- 9 =10

whole number is equal to the number of protons.

8 0
4 years ago
Read 2 more answers
29) What type of chemical reaction is given by the following balanced equation: Al2O3 → Al + O2
Law Incorporation [45]

Decomposition.

Decomposition is the chemical reaction in which a compound breaks down into the two elements or molecules that formed the compound. In this case, Dialuminium Trioxide has broken down into the basic components of Al2 and O2.

Hope this helps!

7 0
3 years ago
If 20 grams of zinc reacts with hydrochloric acid, how much zinc chloride is produced?
Oksi-84 [34.3K]
Assuming there is excess acid, the equation for the reaction is Zn + 2HCl -> ZnCl2 + H2.
Mols of zinc = 20/65 (mass/relative atomic mass)
Ratio of Zn : ZnCl2 = 1 : 1
Hence mols of Zinc = mols of Zinc chloride
Mass of ZnCl2 = (20/65)(65+71)
= 41.8g
3 0
4 years ago
A teenager is discovered in his bedroom by his mother when she comes in to wake him for school at 6 a.m. He is stiff and lividit
Kryger [21]

(I hope I am right.)


I think it would be 5 a.m.

Have a lovely day! ~Pooch ♥

4 0
3 years ago
In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
Alexxandr [17]

Answer:

Velocity, u = 14.7 m/s

Explanation:

It is given that, a driver can probably survive an acceleration of 50 g that lasts for less than 30 ms, but in a crash with a 50 g acceleration lasting longer than 30 ms, a driver is unlikely to survive.

Let v is the highest speed that the car could have had such that the driver survived. Using a = -50 g and t = 30 ms

Using first equation of kinematics as :

v=u+at

In case of crash the final speed of the driver is, v = 0

0=u+at

-u=at

-u=-50\times 9.8\times 30\times 10^{-3}

u = 14.7 m/s

So, the highest speed that the car could have had such that the driver survived is 14.7 m/s. Hence, this is the required solution.

8 0
3 years ago
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