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tatiyna
3 years ago
7

A meteor is moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement desc

ribe the meteors speed or velocity?
Chemistry
2 answers:
Radda [10]3 years ago
6 0

Answer:

since a direction was specifically given, that means it is displacement, and displacement is to velocity while distance is to speed

Explanation:

HACTEHA [7]3 years ago
3 0

Answer:

it's describes the velocity. since a direction was specifically given, that means it is displacement, and displacement is to velocity while distance is to speed

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IV.2. The following problem considers the combustion of butane in a torch. Molecular weight of butane 58.0 g/mol. 2C4H10 + 1302
Anarel [89]

Answer:

(a) Oxygen

(b) 0.84 g

(c) 2.54 g

Explanation:

(a)

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For butane

Given mass = 1.00 g

Molar mass of butane = 58.0 g/mol

Moles of butane = 1.00 g / 58.0 g/mol = 0.0172 moles

Given: For O_2

Given mass = 3.00 g

Molar mass of O_2 = 32.0 g/mol

Moles of O_2 = 3.00 g / 32.0 g/mol = 0.09375 moles

According to the given reaction:

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

2 moles of butane react with 13 moles of O_2

1 mole of butane react with 13/2 moles of O_2

0.0172 moles  of butane react with (13/2)*0.0172 moles of O_2

Moles of O_2 required = 0.1118 moles

Available moles of CuSO_4 = 0.09375 moles

<u>Limiting reagent is the one which is present in small amount. Thus, O_2 is limiting reagent. (0.09375 < 0.1118 )</u>

(b)

The formation of the product is governed by the limiting reagent. So,

13 moles of O_2 react with  2 moles of butane

1 mole of O_2 react with  2/13 moles of butane

0.09375 mole of O_2 react with  (2/13)*0.09375  moles of butane

Moles of butane used = 0.0144 moles

Molar mass of butane = 58.0 g/mol

<u>Mass of butane used = Moles × Molar mass = 0.0144 × 58.0 g = 0.84 g</u>

(c)

13 moles of O_2 on reaction forms 8 moles of carbon dioxide

1 mole of O_2 on reaction forms 8/13 moles of carbon dioxide

0.09375 mole of O_2 on reaction forms (8/13)*0.09375 moles of carbon dioxide

Moles of carbon dioxide obtained = 0.05769 moles

Molar mass of CO_2 = 44.0 g/mol

<u>Mass of CO_2 = Moles × Molar mass = 0.05769 × 44.0 g = 2.54 g</u>

6 0
3 years ago
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