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bixtya [17]
3 years ago
8

Lonic Bonding

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
3 0
Na has a +1 charge, and O has a -1 charge.
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Turbines are 100 percent efficient. True False
max2010maxim [7]
False because there’s only 60% turbulence 20% heat 20% sound
6 0
3 years ago
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A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of
Aleksandr-060686 [28]

<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

<u>Explanation:</u>

To calculate the mole percent of a substance, we use the equation:

\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

8 0
3 years ago
Which organism in a food web would be best described as a recycler?
Anika [276]

C,

takes dead things, makes them into other things

7 0
2 years ago
What is the molarity (molar concentration, unit = M) of K+ found in 200 mL 0.2 M K2HPO4 solution?
enyata [817]

Answer:

0.4 M

Explanation:

The process that takes place in an aqueous K₂HPO₄ solution is:

  • K₂HPO₄ → 2K⁺ + HPO₄⁻²

First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:

  • 200 mL * 0.2 M = 40 mmol K₂HPO₄

Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:

  • 40 mmol K₂HPO₄ * \frac{2mmolK^+}{1mmolK_2HPO_4} = 80 mmol K⁺

Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:

  • 80 mmol K⁺ / 200 mL = 0.4 M
5 0
2 years ago
What is my theoretical yield (in moles) of Potassium Bromide (KBr) if I start with 40 grams of Iron (II) Bromide [FeBr2]? moles
Verizon [17]
The reaction will be: FeBr2 + K --> KBr + Fe
Balancing gives: FeBr2 + 2K --> 2KBr + Fe
The molar mass of FeBr2 is 55.85 + 2*79.9 = 215.65 g/mol.
We divide 40 g / 215.65 g/mol = 0.185 mol FeBr2
Based on stoichiometry:
(0.185 mol FeBr2)(2 mol KBr/1 mol FeBr2) = 0.370 mol KBr
4 0
3 years ago
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