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3 years ago

During an experiment, Sandy recorded incorrect observations and measurements. What would be the most

Chemistry

1 answer:

ruslelena [56]3 years ago

3 0

Sandy would reach an incorrect outcome.that's because she will repeat this mistake in the future too.

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You are on an alien planet where the names for substances and the units of measures are very unfamiliar. Nonetheless, you obtain

Kaylis [27]

The lightbulb contains 40 hours.

3 0

3 years ago

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white

Alex_Xolod [135]

The pH = 2.41

<h3>Further explanation</h3>

Given

5.0% by mass solution of acetic acid

the density of white vinegar is 1.007 g/cm3

Required

Solution

Molarity of solution :

$\tt M=\dfrac{\%mass\times \rho\times 10}{MW~acetic~acid}\\\\M=\dfrac{5\times 1.007\times 10}{60}\\\\M=0.839$

Ka for acetic acid = 1.8 x 10⁻⁵

[H⁺] for weak acid :

$\tt [H^+]=\sqrt{Ka.M}$

Input the value :

$\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41$

7 0

3 years ago

Gaseous methane (CH4) reacts with gaseous oxygen gas (02) to produce gaseous carbon dioxide (CO2) and gaseous water (H20). What

Mariana [72]

Answer:

Theoretical yield = 3.51 g

Explanation:

The formula for the calculation of moles is shown below:

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

For $CH_4$ :-

Mass of $CH_4$ = 1.28 g

Molar mass of $CH_4$ = 16.04 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{1.28\ g}{16.04\ g/mol}$

$Moles_{CH_4}= 0.0798\ mol$

For $O_2$ :-

Mass of $O_2$ = 10.1 g

Molar mass of $O_2$ = 31.998 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$Moles= \frac{10.1\ g}{31.998\ g/mol}$

$Moles_{O_2}= 0.3156\ mol$

According to the given reaction:

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

1 mole of methane gas reacts with 2 moles of oxygen gas

0.0798 mole of methane gas reacts with 2*0.0798 moles of oxygen gas

Moles of oxygen gas = 0.1596 moles

Available moles of oxygen gas = 0.3156 moles

<u>Limiting reagent is the one which is present in small amount. Thus, $CH_4$ is limiting reagent. </u>

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

1 mole of methane gas on reaction produces 1 mole of carbon dioxide.

0.0798 mole of methane gas on reaction produces 0.0798 mole of carbon dioxide.

Mole of carbon dioxide = 0.0798 mole

Molar mass of carbon dioxide = 44.01 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

$0.0798\ moles= \frac{Mass}{44.01\ g/mol}$

Mass of $CO_2$ = 3.51 g

<u> Theoretical yield = 3.51 g</u>

3 0

3 years ago

Which is an example of how people use nonrenewable energy resources?

Arturiano [62]

Answer:

i dont think its walking to school because that doesn't use any resources.

Explanation:

i think its driving a car because we use gasoline for that and for sailing we use the wind. hopefully that helps.

7 0

2 years ago

How many moles of lithium nitrate<br> (LINO3) are in 256 mL of a 0.855 M<br> solution?

sweet [91]

Answer:

There are 0.219 mol of LINO3

7 0

3 years ago