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elixir [45]
3 years ago
12

Unit Test Unit Test Active 13 TIME REMAINING 01:31:26 Chemical A and Chemical B react in an exothermic reaction. What can be kno

wn about what will happen when Chemical A and Chemical B are mixed together? O The new substance will need more energy to form its chemical bonds than the old substance will release. More energy will be released from the old substance than the new substance will need to form its chemical bonds. The color will change as a result of the reaction. O The substance will bubble as a result of the reaction. Save and Exit Next Submit Mark this and return​
Chemistry
2 answers:
Anastasy [175]3 years ago
5 0

Answer:

More energy is released from the old substance than the new substance needs to form its chemical bonds

Explanation:

Edge 2020

Ber [7]3 years ago
4 0

Answer:

More energy is released from the old substance than the new substance needs to form its chemical bonds

Explanation:

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If an atom of the following element had a neutral charge, how many electrons would it have?
Yakvenalex [24]

Answer:

31

Explanation:

In order for an atom to have a neutral charge, it needs to have the same number of protons and electrons. You can see by the illustration that the atomic number is 31, so this element has 31 protons. Knowing that, we also know that it must have 31 electrons to be neutral.

8 0
2 years ago
How many grams of oxygen is needed to completely react with 9.30 moles of aluminum
Orlov [11]
Balanced chemical equation:

* moles of oxygen

4 Al + 3 O2 = 2 Al2O3

4 moles Al -------------- 3 moles O2
9.30 moles Al ---------- moles O2

moles O2 = 9.30 * 3 / 4

moles O2 = 27.9 / 4 => 6.975 moles of O2

Therefore:

Molar mass O2 = 31.9988 g/mol

n = m / mm

6.975 = m / 31.9988

m = 6.975 * 31.9988

m = 223.19 of O2


5 0
3 years ago
What is a change in the arrangement of bases in a gene called?????
Jobisdone [24]
Asexual reproduction.<span>
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4 0
3 years ago
Read 2 more answers
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Explanation:

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6 0
2 years ago
Ammonia, NH3 is a common base with Kb of 1.8 X 10-5. For a solution of 0.150 M NH3:
Vesnalui [34]

The concentrations : 0.15 M

pH=11.21

<h3>Further explanation</h3>

The ionization of ammonia in water :

NH₃+H₂O⇒NH₄OH

NH₃+H₂O⇒NH₄⁺ + OH⁻

The concentrations of all species present in the solution = 0.15 M

Kb=1.8 x 10⁻⁵

M=0.15

\tt [OH^-]=\sqrt{Kb.M}\\\\(OH^-]=\sqrt{1.8\times 10^{-5}\times 0.15}\\\\(OH^-]=\sqrt{2.7\times 10^{-6}}=1.64\times 10^{-3}

\tt pOH=-log[OH^-]\\\\pOH=3-log~1.64=2.79\\\\pH=14-2.79=11.21

8 0
2 years ago
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