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faltersainse [42]
3 years ago
10

How man atoms of Mg are in the following equation: (Mg3 N2)4

Chemistry
2 answers:
mariarad [96]3 years ago
3 0

there are 12 atoms of mg in the following equations

sergey [27]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 12 \ Mg \ atoms }}

Explanation:

We are given this formula:

(Mg_3N_2)_4

We want to find the number of magnesium atoms. We can see that there is a subscript of 3 after the magnesium, so there are 3 magnesium atoms in 1 molecule of the compound.

However, there is a subscript of 4 that comes after the entire compound, so there are actually 4 molecules, with each molecule containing 3 magnesium atoms. Multiply 4 and 3.

  • 3 Mg * 4 = 12 Mg

There are 12 magnesium atoms.

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Indicate whether the following are acidic, basic, or neutral solution
goldenfox [79]
You answer this by using the pH formula and and the relation of pH and pOH, pH = -log[H+] and 14 = pH + pOH. The correct classification are as follows:

<span>A. [H2O+]=6.0x10^-12
basic

B. [H3O+]=1.4x10^-9
basic

C. [OH-]=5.0x10^-12
acidic

D. {OH-]=3.5x10^-10
acidic

Hope this answers the question.

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5 0
3 years ago
The energy transfer between the system and the surroundings, in the form of heat at constant pressure is described by which of t
kicyunya [14]
I think the answer would be d but not 100% sure
4 0
3 years ago
What type of reaction is BaI^2+H^2SO^4 2H+BaSO^4
anyanavicka [17]

Answer:  It is a Double Displacement(Replacement) Reaction.

Explanation:

6 0
3 years ago
What is the ph of a solution that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate. (ka for propanoic acid = 1.3 x 10
max2010maxim [7]
<span>The pH is given by the Henderson - Hasselbalch equation:
              pH = pKa + log([A-]/[HA])
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3 0
3 years ago
One of the radioactive isotopes used in medical treatment or analysis is chromium-51. The half-life of chromium-51 is 28 days. H
Goryan [66]

Answer : The time required for decay is, 84 days.

Explanation :

Half-life of chromium-51 = 28 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{28\text{ days}}

k=0.0248\text{ days}^{-1}

Now we have to calculate the time required for decay.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = time taken by sample = ?

a = let initial activity of the sample = 100

a - x = amount left after decay process  = 12.5

Now put all the given values in above equation, we get

t=\frac{2.303}{0.0248}\log\frac{100}{12.5}

t=83.9\text{ days}\approx 84\text{ days}

Therefore, the time required for decay is, 84 days.

7 0
3 years ago
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