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2 years ago

Explain the difference between a front and an air mass.

1 answer:

8 0

Answer: The differences is Air masses cover over thousands and hundreds and millions of square kilo. A front is a boundary which two air masses but different temperature and moisture content meet.

Explanation:

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100 points! please help me will mark brainliest

aalyn [17]

S + O2 → SO2

Moles of S:

11/32.06 = 0.3431 mol

Moles of O2:

44/15.999 = 2.75 mol

Sulfur will be the limiting reagent.

Oxygen will be the excess reagent

2.75 - 0.3431 = 2.4071 mols of oxygen leftover

2.4071 * 15.999 = 38.511

38.511 grams of oxygen will be leftover

7 0

2 years ago

Read 2 more answers

Please help!!

deff fn [24]

Im answering question 2
Because it increases the kinetic energy of reactants and so the frequency of collision between reactants. Thus, provided the temperature is not too high to lead to denaturation of reactants, increase in temperature increases the rate of reaction

And for ques. 7:
It gets affected when gaseous reactants in which products formed have less volume than reactants, so increase in pressure increases the rate of reaction

4 0

3 years ago

Question 1(Multiple Choice Worth 4 points)

zubka84 [21]

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives → 3 × 0.344 moles of CO

= 1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$