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kupik [55]
2 years ago
7

50 points for help. Refer to the observations of the test tubes from part A. Determine which metal (or hydrogen) in each test tu

be is more reactive. Remember that the less reactive metal (or hydrogen) will typically end up in pure form as an element, so no reaction will occur if the less reactive metal (or hydrogen) is the one that begins in pure form.

Chemistry
1 answer:
Andrei [34K]2 years ago
8 0

Metals that are higher in the activity series are more reactive than metals that are lower in the activity series.

<h3>Reactivity of metals</h3>

The reactivity of metals is determined by the position of the metals in the  activity series of metals. Metals that are more reactive are found at the top of the series while metals that are less reactive are found at the bottom of the series.

In each case, the more reactive metal is as follows;

  1. Iron is more reactive
  2. Hydrogen is more reactive
  3. Zinc is more reactive
  4. Magnesium is more reactive
  5. Zinc is more reactive
  6. Iron is more reactive

Learn more about reactivity series of metals:brainly.com/question/24866635

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Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C
Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, E = \triangle M * 931.5

\triangle M = \sum M_{product} - \sum M_{reactant}

Mass of 1H, M_{H} = 1.007823

Mass of 2H, M_{2H} = 2.0141u

Mass of 3H, M_{3H} = 3.016 u

Mass of Helium, M_{4He} = 4.002602u

Mass of Beryllium, M_{7Be} = 7.01693 u

Mass of neutron, M_{n} = 1.008664 u

a) 2H + 3H \rightarrow 4He + n

\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u

Energy released,

E = -0.01884 * 931.5\\E = -17.55 Mev

Energy released = 17.55 MeV

b) 4He + 4He \rightarrow 7Be + n

\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} +  M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u

Energy released,

E = 0.02039 * 931.5\\E = 18.99 Mev

c) 2H + 2H \rightarrow 3 He + n

\triangle M = (M_{3He} + M_{n} ) - (M_{2H} +  M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u

Energy released,

E = -0.003536 * 931.5\\E = -3.29 Mev

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) 2H + 2H \rightarrow 3H + 1H

\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} +  M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u

E = -0.00435 * 931.5\\E =-4.075 Mev

E = 4.075 MeV ( Energy is released)

5 0
3 years ago
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Novay_Z [31]

Answer:

3rd or 4th answer is correct

8 0
3 years ago
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How can valence electrons be used to predict the type of compounds formed?
Alex17521 [72]
Valence electrons are the electrons in the outermost shell. Those are t<span>he electrons on an atom that can be gained or lost in a chemical reaction.
</span>Elements that are left on the periodic table <span> have relatively few </span>valence electrons<span>, and can form ions  more easily by losing their </span>valence electrons<span> to form positively charged cations.</span>
<span>Nonmetals are further to the right on the periodic table, so they gain electrons relatively easily and lose them with difficulty. </span>

7 0
3 years ago
What is constant on wet asphalt?
wariber [46]

Answer:

Friction and Automobile Tires. ... On dry surfaces you might get as high as 0.9 as a coefficient of friction, but driving them on wet roads would be dangerous since the wet road coefficient might be as low as 0.1

Explanation:

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3 years ago
If 125 grams of Cu reacts 75 grams of HNO3, how many grams of Cu(NO3) form?
melisa1 [442]

Answer: 37.5grams of Cu(NO3)2

Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2

<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>

<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>

<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>

<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>

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3 years ago
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