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2 years ago

What is the Empirical Formula when the Molecular Formula is Sb17Br21

1 answer:

5 0

Difference Between Empirical and Molecular Formulas. The key difference between empirical and molecular formulas is that an empirical formula only gives the simplest ratio of atoms whereas a molecular formula gives the exact number of each atom in a molecule. In chemistry, we often use symbols to identify elements and molecules.

Hope I helped :)

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Assuming equal concentrations, rank these solutions by ph.? assuming equal concentrations, rank these solutions by ph. koh (aq h

polet [3.4K]

I remember coming across this question and the options were:
KOH, HCN, NH₃, HI, Sr(OH)₂

Now, a substance with a low pH is one that dissociates completely in water to release hydrogen ions, while basic substances dissociate completely to release hydroxide ions. Therefore, in the order of increasing pH:
HI, HCN, NH₃, Sr(OH)₂, KOH

7 0

2 years ago

Ionic bonds are strong; therefore, ionic compounds

Bess [88]

This is not a question
what are you asking

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2 years ago

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Tia has a sample of pure gold (Au). She weighed the sample and the result was 35.9 grams. Tia wants to determine the number of a

Gemiola [76]

First, find moles of gold given the mass of the sample:
(35.9g Au)/(197.0g/mol Au) = 0.182mol Au

Second, multiply moles of Au by Avogrado's number:
(0.182mol)(6.02 x10^23)= 1.10x10^23 atoms Au

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3 years ago

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You are carrying out an experiment in the lab to study the glycolysis pathway. To do this, a liver extract (which is capable of

Virty [35]

Answer:

When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.

Aldolase cleaves a hexose into two trioses.

[See the image attached].

Asterisk indicates the label.

When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.

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2 years ago

A 0.885 M solution of KBr whose initial volume is 82.5 mL has more water added until its concentration is 0.500 M. What is the n

4vir4ik [10]

Answer:

$V_2=146mL$

Explanation:

Hello there!

In this case, since the equation for the calculation of dilutions is:

$M_1V_1=M_2V_2$

Whereas M is the molarity and V the volume, because the final concentration is lower than the initial. Thus, since we are asked to calculate the final volume, we solve for V2 as follows:

$V_2=\frac{M_1V_1}{M_2}=\frac{0.885M*82.5mL}{0.500M}\\\\V_2=146mL$

Best regards!

4 0

2 years ago