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ololo11 [35]
3 years ago
10

In an endothermic reaction would energy be considered a reactant or a product?

Chemistry
1 answer:
timurjin [86]3 years ago
4 0

Answer:

In an endothermic reaction, the products have more stored chemical energy than the reactants. In an exothermic reaction, the opposite is true. The products have less stored chemical energy than the reactants. The excess energy in the reactants is released to the surroundings

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Someone please help me and answer my question asap. i need to turn my assignment in soon... please.
Ksenya-84 [330]

Answer:

Explanation:

a. the salt produced would be Mg3N2(magnesium nitride)

b. magnesium loses 2 electron to form Mg2+ ion and nitrogen gains 3 electron to form n3-

when several of these ions come together 3 Mg2+ ion combine with 2 n3- ion to form Mg3N2 thus Mg getting six electron from nitrogen to form a ionic bond.

c. the reaction is not balanced Mg + N2 = Mg3n2

to make it balanced the reaction should be 3 Mg + N2 = Mg3N2.

the reaction was not balanced before because the number of Mg on both side of the reaction was not equal.

d. magnesium nitrate has formula Mg(NO3)2 is formed when Mg combines with nitrogen and oxygen Mg + N2 + o2

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3 years ago
What would be the atomic mass of Nitrogen-14​
Maurinko [17]

Answer:

The atomic mass of nitrogen is

<h2>14.0067 u</h2>

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7 0
3 years ago
Read 2 more answers
How do I calculate the standard enthalpy change for the following reaction at 25 °C ?
ladessa [460]
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g) 

Using the standard enthalpies of formation given in the source below: 

(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ 
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MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
4 0
4 years ago
A 10 mm Brinell hardness indenter is used for some hardness testing measurements of a steel alloy. a) Compute the HB of this mat
Lady_Fox [76]

Explanation:

(a)  The given data is as follows.  

         Load applied (P) = 1000 kg

         Indentation produced (d) = 2.50 mm

         BHI diameter (D) = 10 mm

Expression for Brinell Hardness is as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}    

Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

                       = \frac{2000}{9.98}                          

                       = 200

Therefore, the Brinell HArdness is 200.

(b)     The given data is as follows.

               Brinell Hardness = 300

                Load (P) = 500 kg

               BHI diameter (D) = 10 mm

             Indentation produced (d) = ?

                      d = \sqrt{(D^{2} - [D - \frac{2P}{HB} \pi D]^{2})}

                         = \sqrt{(10 mm)^{2} - [10 mm - \frac{2 \times 500 kg}{300 \times 3.14 \times 10 mm}]^{2}}

                          = 4.46 mm

Hence, the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used is 4.46 mm.

5 0
3 years ago
1 -Is Apple Juice with No Pulp a pure substance or a mixture ?
-Dominant- [34]

1. Pure substance

2.mixture

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3 years ago
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