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anygoal [31]
3 years ago
5

Which subatomic particle is involved in chemical bonding and is responsible for an element's reactivity?

Chemistry
2 answers:
Murrr4er [49]3 years ago
6 0
<span>There are three different subatomic particles present in the atoms of each element: neutron, proton and </span>electron<span>. It is the </span>electrons<span>, and more specifically the valence </span>electrons<span>, that determine the reactivity of an element.</span>
maxonik [38]3 years ago
3 0
The answer is electrons
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What is the net force acting on a 3000g sled that accelerated at 5 m/s2 ?
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Answer:

<h2>15 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

3000 g = 3 kg

We have

net force = 3 × 5 = 15

We have the final answer as

<h3>15 N</h3>

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Which expression represents a reaction rate?a. time/massc. energy/timeb. number/timed. time/energy
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The right answer for the question that is being asked and shown above is that: "<span>b. number/timed." Reaction Rate refers to the </span> speed of reaction<span> for a reactant or product in a particular </span>reaction<span> is intuitively defined as how fast or slow a</span>re action<span> takes place.</span>
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What are "representative elements"?
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How can you described the energy of a storm
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4 0
3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
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