What is the volume of 120mg of water

Consider the reaction between reactants s and o2: 2s(s)+3o2(g)→2so3(g)if a reaction vessel initially contains 7 mols and 9 mol o

nadya68 [22]

Answer: -

1 mol

Explanation: -

Number of moles of Sulphur S = 7

Number of moles of O2 = 9

The balanced chemical equation for the reaction is

2S (s)+3 O2 (g)→2SO3(g)

From the above reaction we can see that

3 mol of O2 react with 2 mol of S

9 mol of O2 will react with $\frac{2 mol S x 9 mol O2}{3 mol O2}$

= 6 mol of S

Unreacted S = 7 - = 1 mol.

If a reaction vessel initially contains 7 mol S and 9 mol O2

1 mole of s will be in the reaction vessel once the reactants have reacted as much as possible

6 0

3 years ago

For the following reaction, identify whether the compound in bold is behaving as an acid or a base.

harina [27]

Answer: acid

Explanation: it dissolves in water to produce hydronium ion as the only positive ion

3 0

3 years ago

A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How

IrinaK [193]

Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $783J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 254 g

$\Delta T$ = change in temperature = $T_2-T_1=(23.73-26.01)=-2.28^oC$

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

$q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]$

$q=-4208.28J=-4.81kJ$

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

7 0

3 years ago

AB G C F G Y OBLAD

viktelen [127]

Blade resistance hair motion pull down

4 0

2 years ago

A dark brown binary compound contains oxygen and a metal. It is 13.38% oxygen by mass. Heating it moderately drives off some of

Leto [7]

Answer:

a) Mass of O in compound A = 32.72 g

Mass of O in compound B = 21.26 g

Mass of O in compound C = 15.94 g

b) Compound A = MO2

Compound B = M3O4

Compound C = MO

c) M = Pb

Explanation:

Step 1: Data given

A binairy compound contains oxygen (O) and metal (M)

⇒ 13.38 % O

⇒ 100 - 13.38 = 86.62 % M

After heating we get another binairy compound

⇒ 9.334 % O

⇒ 100 - 9.334 = 90.666 % M

After heating we get another binairy compound

⇒ 7.168 % O

⇒ 100 - 7.168 = 92.832 % M

The first compound has an empirical formula of MO2

⇒ 1 mol M for 2 moles O

Step 2: Calculate amount of metal and oxygen in each

compound A: M = m1 *0.8662 O = m1 *0.1338

compound B: M = m2 *0.90666 O = m2 *0.09334

compound C: M = m3 *0.92832 O = m3 *0.07168

Step 3: Calculate mass of oxygen with 1.000 grams of M

Compound A: 1.000g * 0.1338 m1gO / 0.8662m1gMetal = 0.1545

Compound B: 1.000g * 0.09334 m2gO / 0.90666m2gMetal = 0.1029

Compound C: 1.000g * 0.07168 m3gO / 0.92832m3gMetal = 0.07721

Step 4:

1 mol MO2 has 1 mol M and 2 moles O

m1 = (mol O * 16)/0.1338 m1 = 239.2 grams

1 mol M = 0.8632*239.2 = 206.48

0.90666m2 = 206.48 ⇒ m2 = 227.74 g

0.92832m3 = 206.48 ⇒ m3 = 222.42 g

Step 5: Calculate mass of O

Mass of O in compound A = 239.2 - 206.48 = 32.72 g

Mass of O in compound B = 227.74 - 206.48 = 21.26 g

Mass of O in compound C = 222.42- 206.48 = 15.94 g

Step 6: Calculate moles

Moles of O in compound A ≈ 2

⇒ MO2

Moles of O in compound B = 21.26 / 16 ≈ 1.33

⇒ M3O4

Moles of O compound C = 15.94 /16 ≈ 1 moles

⇒ MO

Step 7: Calculate molar mass

The mass of 1 mol metal is 206.48 grams ⇒ molar mass ≈ 206.48 g/mol

The closest metal to this molar mass is lead (Pb)

6 0

2 years ago