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1 year ago

What distance in kilometers can you drive if your car gets 14.00 kilometers per liter and you have 35.00 liters of gas?

Chemistry

1 answer:

agasfer [191]1 year ago

7 0

2.5 km is the distance you can travel if a car gets 14 kilometers per liter and have 35 liters of gas.

Total distance traveled in 1 liter = 14 km

Total gas in the car = 35 liters

We have to find how much distance can be covered in in35 liters of gas.

Distance is the length of the path traveled between two points. It is the total movement of an object in a direction. It measures length.

Distance = 35/14

Or, distance = 2.5 km

Therefore, the total distance that can be covered by a car in 35 liters of gas is 2.5 km.

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A compound contains 74.2 g Na and 25.8 g O. Determine the Empirical Formula for this compound

Ksenya-84 [330]

Answer:

The empirical formula for the compound is Na2O

Explanation:

Data obtained from the question include:

Sodium (Na) = 74.2g

Oxygen (O) = 25.8g

We can obtain the empirical formula for the compound as follow:

First, divide the above by their individual molar mass as shown below:

Na = 74.2/23 = 3.226

O = 25.8/16 = 1.613

Next, divide the above by the smallest number

Na = 3.226/1.613 = 2

O = 1.613/1.613 = 1

Therefore, the empirical formula is:

Na2O

4 0

3 years ago

Read 2 more answers

(a.) a 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g).

Natali [406]

The individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

Moles of CO2 formed can be calculated as

= Mass of CO2 / Molar mass of CO2

= 1.9061 / 44 = 0.0433 moles

<h3>Calculation of no. of moles of carbon</h3>

Now, moles of C which is present in one mole of CO2 = 1 mole

Moles of C in 0.0433 moles of CO2 = 0.0433 moles

As we know that, molar mass of C = 12 g / mol

Mass of C in 0.7549 g of given sample can be calculated as

= 0.0433 × 12 =0.5196 g

Mass of H2O formed = 0.3370 g

Similarly, Molar Mass of H2O = 18 g / mol

Moles of H2O = 0.3370 / 18 = 0.0187 moles

Moles of H present in 1 mole of H2O = 2 moles

Moles of H present in 0.0187 mole of H2O = 2 × 0.0187 = 0.0374 moles

Molar mass of H = 1 g / mol

Mass of H contained in 0.7549 g of sample = 1 × 0.0374= 0.0374 g

Mass of O in 0.7549 g sample can be calculated as

= 0.7549 – [(Mass of C ) + (Mass of H) ]

= 0.7549 – [ (0.5196) + (0.0374) ]

= 0.1979 g

Thus, we calculated that the individual mass of C, H and O in given sample are 0.5196 g, 0.0374 g and 0.1979 g respectively.

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DISCLAIMER: THE above question is incomplete. Complete question is given below:

A 0.7549g sample of the compound burns in o2(g) to produce 1.9061g of co2(g) and 0.3370g of h2o(g). Calculate the individual mass of C, H and O in the given sample.

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2 years ago

WWW<br> 7. What is the subscript for Hydrogen?

Korolek [52]

Answer: In the chemical formula for water, the subscript for hydrogen is 2. Notice that the 2 is smaller and written slightly below the H and O. It is called a subscript because it is written ("script") "below" ("sub") the preceding letter.

Explanation:

4 0

3 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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4 years ago

What is the SI unit for volume?

kirill115 [55]

The SI unit for volume is cubic meters of M^3

4 0

3 years ago

Read 2 more answers

What distance in kilometers can you drive if your car gets 14.00 kilometers per liter and you have 35.00 liters of gas?​

What distance in kilometers can you drive if your car gets 14.00 kilometers per liter and you have 35.00 liters of gas?