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3 years ago

11

Which element has an atomic number of 16? a. oxygen (O) b. sulfur (S) c. germanium (Ge)

2 answers:

natta225 [31]3 years ago

6 0

Answer will be B : sulfur

emmasim [6.3K]3 years ago

3 0

Answer: Sulfur

Explanation:

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1. Related to the number of particles in a gram of

Romashka-Z-Leto [24]

Answer:

Avogadro's number or Avogardro’s constant

Explanation:

I’m pretty sure this is correct if it’s not I’m sorry lol.

6 0

2 years ago

Why do most atoms form chemical bonds?

zloy xaker [14]

Answer:

Why do most atoms form chemical bonds? They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable. Double and triple covalent bonds that have greater bond energy and are shorter than single bonds.

Explanation: HOPE THIS HELPS YOU..

3 0

3 years ago

A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

4 years ago

Read 2 more answers

Calcula el pH de estas soluciones: A-acido sulfúrico 0,0013 m B- hidróxido de socio 1,7 . 10 (-3) C- hidróxido de bario 2,2 . 10

Vladimir [108]

Answer:

ksüqkdbkavavsü hwkfülbjvvsb

4 0

3 years ago

What is the molarity of a KF(aq) solution containing 3.0 mol of KF in 2.00L of solution?

sergeinik [125]

Answer : The molarity of $KF$ in the solution is 1.5 M.

Explanation : Given,

Moles of $KF$ = 3.0 mol

Volume of solution = 2.00 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Moles of }KF}{\text{Volume of solution (in L)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{3.0mol}{2.00L}=1.5mole/L=1.5M$

Therefore, the molarity of $KF$ in the solution is 1.5 M.

4 0

3 years ago