Question 3 someone help me please

How many formula units are in 4.52 moles of H3SO3? Type your answer

lakkis [162]

Answer:

The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.

Quick conversion chart of moles H2SO3 to grams

1 moles H2SO3 to grams = 82.07908 grams

2 moles H2SO3 to grams = 164.15816 grams

3 moles H2SO3 to grams = 246.23724 grams

4 moles H2SO3 to grams = 328.31632 grams

5 moles H2SO3 to grams = 410.3954 grams

6 moles H2SO3 to grams = 492.47448 grams

7 moles H2SO3 to grams = 574.55356 grams

8 moles H2SO3 to grams = 656.63264 grams

9 moles H2SO3 to grams = 738.71172 grams

10 moles H2SO3 to grams = 820.7908 grams

7 0

3 years ago

Suppose a 500.mL flask is filled with 1.0mol of CO, 1.5mol of H2O and 0.70mol of CO2. The following reaction becomes possible: +

valentina_108 [34]

Answer:

[CO] = 0.62 M

Explanation:

Step 1: Data given

Volume of the flask = 500 mL

Number of moles CO = 1.0 moles

Number of moles H2O = 1.5 moles

Number of moles CO2 = 0.70 moles

The equilibrium constant K for this reaction is 3.80

Step 2: The balanced equation

CO(g) + H2O(g) ⇆ CO2(g) +H2(g)

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CO] = 1.0 moles / 0.500 L = 2.0 M

[H2O] = 1.5 moles / 0.500 L = 3.0 M

[CO2] = 0.70 moles / 0.500 L = 1.4 M

[H2] = 0M

Step 4: The concentration at the equilibrium

For 1 mol CO we have 1 mol H2O to produce 1 mol CO2 and 1 mol H2

[CO] = 2.0 -X M

[H2O] = 3.0 - X M

[CO2] =1.4 + X M

[H2] = X M

Step 5: Define Kc

Kc = [CO2][H2]/ [CO][H2O]

3.80 = (1.4 + X) * X / ((2.0 - X)*3.0 -X))

X = 1.38

[CO] = 2.0 -1.38 = 0.62 M

[H2O] = 3.0 - 1.38 = 1.62 M

[CO2] =1.4 + 1.38 M = 2.78

[H2] = 1.38 M

Kc = (2.78*1.38) / (0.62*1.62)

Kc = 3.8

[CO] = 0.62 M

8 0

3 years ago

Which of the following statements describes a solid?

Andrew [12]

B.) Its particles are packed together tightly.

4 0

3 years ago

Read 2 more answers

How do you write the balanced equation for strontium hydroxide and zirconium (I) perchlorate

Andrei [34K]

The law of conservation of mass states that the mass of a system must remain constant everytime, it can neither be created or destroyed. This means the number of atoms of each element on reactant side must equal the number of atoms of each element on product side.

First we figure out the chemical formula for each compound by taking advantage of oxidation states which are able to tell us whether an element will lose or gain an electron

Strontium hydroxide?

Sr is an alkaline earth metal which has 2 valence electrons that when both are lost it has an oxidation state of +2. Since we know the hydroxide ion has a charge of -1, $-_{OH}$ , Then two $-_{OH}$ are needed to form compound, hence we have $SrOH_2$

zirconium (I) perchlorate?

The oxidation state of zirconium is +1 since it is stated that we have zirconium(I). Converserly, perchlorate has a -1 charge which further proves that zirconium has a +1 oxidation state

The product would have a strontium diperchlorate because we know from earlier that Sr has a +2 oxidation state, and we would also have ZrOH because Zr has a +1 oxidation state and $-_{OH}$ has a -1 charge

The unbalanced equation with all chemicals formulas would be

$SrOH_2_(s) + ZrClO^{-} _4(aq) ==> Sr(ClO^{-} _4)_2(aq) + ZrOH(aq)$

Using the law of conservation of mass, it is clear that there is 1 molecule of $ClO^{-} _4$ on reactant side as compared to 2 molecules of $ClO^{-} _4$ on product side. On the other hand, the are 2 molecules of $-_{OH}$ on reactant side and 1 molecule of $-_{OH}$ on product side. To balance we add a coefficient of 2 on $ZrClO_4$ and a coefficient of 2 on ZrOH,