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KengaRu [80]
3 years ago
9

What is the initial reactant for a fusion reaction in the sun? a) Iron-56 b) Hydrogen-1

Chemistry
1 answer:
jarptica [38.1K]3 years ago
6 0

Hydrogen

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Which enzyme will be most involved in the digestion of high protein food?
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The correct answer should be A

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Which particle may be gained, lost, or shared by an atom when it forms a chemical bond?
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Electrons because the amount of valence electrons determines the bonds it can form and often times during a chemical bond or the forming of a compound an element will lose some of its electrons.

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1. You are in a car at a stop sign. The car accelerates from 0 to 45 mph in 4 seconds. What
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c

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c

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Write the IUPAC and common names, if any, of the carboxylate salts produced in the reaction of each of the following carboxylic
Tems11 [23]

Answer:

Following are the explanation to this question:

Explanation:

The salts of carboxylate are named by the writing, which is also named as the creation of the first, which is followed by the name of the carboxylic acid were '-ic' of the acid end and replaced by the 'ate'.

Following are the description of the given reaction:

In reaction A:

2-Bromopropanoic acid= C_3H_5BrO_2

C_3H_5BrO_2+NaOH⇄ C_3H_4BrNaO_2 +H_2O

The IUPAC name is Sodium-2-Bromopropanate

In reaction B:

2-Methylhexanoic acid= C_7H_{14}O_2

C_7H_{14}O_2+NaOH⇄ C_7H_{13}NaO_{2}+H_2O

The IUPAC name is Sodium-2-Methyl hexanoate

5 0
3 years ago
How man grams of cl2 are consumed to produce 12.0 g of KCl
Korvikt [17]

Answer:

5.71 g

Explanation:

Step 1: Write the balanced equation

2 K + Cl₂ ⇒ 2 KCl

Step 2: Calculate the moles corresponding to 12.0 g of KCl

The molar mass of KCl is 74.55 g/mol.

12.0 g × 1 mol/74.55 g = 0.161 mol

Step 3: Calculate the moles of Cl₂ needed to produce 0.161 moles of KCl

The molar ratio of Cl₂ to KCl is 1:2. The moles of Cl₂ needed are 1/2 × 0.161 mol = 0.0805 mol

Step 4: Calculate the mass corresponding to 0.0805 moles of Cl₂

The molar mass of Cl₂ is 70.91 g/mol.

0.0805 mol × 70.91 g/mol = 5.71 g

8 0
3 years ago
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