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Aliun [14]
3 years ago
9

HELP ASAP THX!!!!!!!!!!!!

Chemistry
2 answers:
Sedbober [7]3 years ago
5 0

The answer would be B, flapping ability.

Ann [662]3 years ago
5 0

B, flapping ability.

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by mixing red and yellow

Explanation:

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How do I find the moles of OH- which reacted (mol) in the titration. Table Attached
BaLLatris [955]

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It is equal to the number of moles of acid that reacted. When Oxalic acid is your limiting reactant it is the # of moles of oxalic acid used. When NaOH is your limiting reactant it is equal to the number of moles of NaOH used.

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Which of the following most accurately describes a regulatory molecule that must be transported to its place of action within th
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2 years ago
If 10.57 g of magnesium reacts completely with 6.96 g of oxygen, what is the percent by mass of oxygen in magnesium oxide? Round
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Answer:

39.7 %

Explanation:

magnesium + oxygen ⟶ magnesium oxide

   10.57 g         6.96 g               17.53 g

According to the <em>Law of Conservation of Mass</em>, the mass of the product must equal the total mass of the reactants.

Mass of MgO = 10.57 + 6.96

Mass of MgO = 17.53 g

The formula for mass percent is

% by mass = Mass of component/Total mass × 100 %

In this case,

% O = mass of O/mass of MgO × 100 %

Mass of O = 6.96 g

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5 0
3 years ago
In an ion with an unknown charge, the total mass of all the electrons was determined to be 2.19 ✕ 10−26 g, while the total mass
Ilya [14]

Answer:

\large \boxed{\text{Fe$^{{2+}}$}}

Explanation:

1. Number of electrons

\text{Number of electrons} = 2.19 \times 10^{-26}\text{ g} \times \dfrac{\text{1 electron}}{9.109 \times 10^{-28}\text{ g}} = \text{24 electrons}

2. Number of protons

\text{Number of protons} = 4.34 \times 10^{-23}\text{ g} \times \dfrac{\text{1 proton}}{1.673 \times 10^{-24}\text{ g}} = \text{26 protons}

3. Identify the ion

An atom with 26 protons is iron, Fe.

A neutral atom of iron would have 26 electrons.

The ion has only 24 electrons, so it has lost two. The ion must have a charge of +2.

\text{The symbol for the ion is $\large \boxed{\textbf{Fe$^{\mathbf{2+}}$}}$}

4 0
3 years ago
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