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3 years ago

What is the oxidation state of sulfur in SO3?

1 answer:

8 0

The oxidation state of a substance is the electric charge it is exhibiting in a given state. This may be determined by looking at the oxidation states of accompanying atoms as well as the charge on the complete molecule.

In this case:

Molecular charge: 0
Oxidation state of oxygen: -2 (it is a group 6 element)

Thus,

S + 3 * -2 = 0
S = 6

Sulfur is exhibiting an oxidation state of +6.

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CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of

neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0

3 years ago

A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$

Now, find moles of $CH_{3} OH$ are needed using the molarity equation:

$M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol$

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M

5 0

3 years ago

_____ NaPO4+_____KOH

Artist 52 [7]

NaPO4 + KOH -> KPO4 + NaOH
already balance

6 0

3 years ago

KMnO4 + KCI + H2SO4<br>MnSO4 + K2S04+ Cl₂ + H₂O

yKpoI14uk [10]

Answer:

Balancing the equation

2KMnO₂+10KCl+8H₂SO₄⇒2MnSO₄+6K₂SO₄+8H₂O+5Cl₂

7 0

3 years ago

química: compara y contrasta la investigación básica, investigación aplicada y el desarrollo tecnológico. razonamiento crítico.

mihalych1998 [28]

This isn’t anything related to chemistry dude

7 0

3 years ago