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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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A beaker holds 962 g of a brine solution that is 6.20 percent salt. If 123g of water are evaporated from the beaker, how much sa
denpristay [2]

Answer:

13.687 grams of salt should be added

The total grams of 8.6% of brine solution produced is 852.687g

Explanation:

Solution mass= 962g

Salt= 6.2%

Water = 93.8%

962 gram of water is made up of:

902.356g ( due to vaporization which reduces mass)

= 902.356 - 123

= 779. 356g of water

59.644g of salt.

If we add x gram of salt for making brine solution up to 8.6%

=(59. 644g + x)g.of salt

% salt = Mass of Salt / Total mass of solution

= 0.086= 59.644 + x / 779.356 + 59.644 + x

= 59.644 + x / 839 + x

x= 13.687 g of salt

Grams of 8.6% brine solution will be:

Gram of water + total gram of salt added to form 8.6% brine solution.

= 779.356g +59.644g + 13.687g

= 852.687g

The total grams of 8.6% of brine solution produced is 852.687g

8 0
3 years ago
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Water (2350 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the ini
Eva8 [605]

Answer:

40.7062 °C  

Explanation:

Let the initial temperature = x °C

Boiling temperature of water = 100 °C

Using,

Q = m C ×ΔT

Where,  

Q is the heat absorbed in the temperature change from x °C to 100 °C.

C gas is the specific heat of the water = 4.184 J/g  °C

m is the mass of water

ΔT = (100 - x) °C  

Given,

Mass = 2350 g

Q = 5.83 × 10⁵ J

Applying the values as:

Q = m C ×ΔT

5.83 × 10⁵ = 2350 × 4.184 × (100 - x)

<u>x, Initial temperature = 40.7062 °C  </u>

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