I NEED THIS NOW NO LINKS OR ILL REPORT PLZZZ

Tameca is planning a hiking trip. The trail she would like to follow is 7.5 miles long. She plans to start her hike at 10:00 am.

V125BC [204]

Answer:

1.5 miles per hour

Explanation:

7.5 miles in 5 hours

7.5/5=1.5 miles per hour

6 0

3 years ago

What is the difference between property and change?

miss Akunina [59]

If your talking about physical property and chemical change then , a physical property is something that can be observed without changing and a chemical change , changes into a new substance.

8 0

3 years ago

Read 2 more answers

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

6 0

3 years ago

How many moles of lithium are present in a sample that contains 2.45x10^87 formula units of Li2SO4?

ddd [48]

The answer is: 8.14·10⁶³ moles of lithium are present.

N(Li₂SO₄) = 2.45·10⁸⁷; number of formula units of lithium sulfate.

n(Li₂SO₄) = N(Li₂SO₄) ÷ Na.

n(Li₂SO₄) = 2.45·10⁸⁷ ÷ 6.022·10²³ 1/mol.

n(Li₂SO₄) = 4.07·10⁶³ mol; amount of lithium sulfate

In one molecule of lithium sulfate, there are two atoms of lithium.

n(Li₂SO₄) : n(Li) = 1 : 2.

n(Li) = 2 · 4.07·10⁶³ mol.

n(Li) = 8.14·10⁶³ mol; amount of lithium atoms.

3 0

3 years ago

Tina's calculations of the tarantula found that the spider was able to cover 20 centimeters in 5 seconds ,what was the average s

12345 [234]

4 centimetres per second

7 0

3 years ago