CHEMISTRY HIGH SCHOOL

Which substance is a reactant(s)
O2 only
MgO only
Mg and O2
Mg only

Answers

Answer 1
Answer: Reactants in all these elements are Mg and O2

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Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.391 g of carbon dioxide is produced from the reaction of 0.16 g of methane and 0.84 g of oxygen gas, calculate the percent yield of carbon dioxide.

Answers

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}=(0.16g)/(16g/mole)=0.01moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(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 (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 (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* \text{ Molar mass of }CO_2

\text{ Mass of }CO_2=(0.013moles)* (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}* 100

\% \text{ yield of }CO_2=(0.391g)/(0.572g)* 100=68.4\%

Therefore, the percent yield of CO_2 is, 68.4 %

What is the percent by mass of water in Na2SO4 • 10H2O?

Answers

We determine the percent by mass of water in the compound by dividing the mass of  water by the total mass. The total mass of Na2SO4.10H2O is equal to 322 g. The mass of water is 180 g. 
       
                    percent by mass of water = (180 / 322)*(100 %) = 55.9%

55.92 percent is the correct answer for online chemistry I just took the test!

The freezing point of bromine is
(1) 539°C (3) 7°C(2) –539°C (4) –7°C

Answers

The freezing point of bromine is –7°C. Therefore, the correct option is option D among all the given options.

What is bromine?

Bromine (Br) is a chemical substance, a deep red toxic liquid, and a member of Group 17 (Group VII) of both the periodic chart. At room temperature, free bromine seems to be a reddish brown fluid with such a significant vapour pressure.

Bromine vapour has an amber color. Bromine has a strong odor and can irritate the skin, eyes, especially respiratory system. Even brief exposure to intense bromine vapour can be lethal. The freezing point of bromine is –7°C.

Therefore, the correct option is option D.

To learn more about bromine, here:

brainly.com/question/1952132

#SPJ2
The freezing point of bromine is (4) -7 degrees Celsius.
Hope this helps~

True or False: Since baking soda is a powder, all the substances it breaks down intoare also powders.

Answers

False; Sodium hydrogen carbonate (also known as sodium bicarbonate or bicarbonate of soda) has the chemical formula NaHCO3. When it is heated above about 80°C it begins to break down, forming sodium carbonate, water and carbon dioxide. This type of reaction is called a thermal decomposition.

Which statement describes an intensive property of matter

Answers

An intensive property is a property that does not depend on the amount of the substance in a system being measured. Examples are density and color. The opposite is called extensive where the properties depends on the amount of a substance.

Suppose that you measure a pen to be 10.5 cm long. Convert this to meters.

Answers

If the pen is 10.5cm long, it is 0.105 meters. This can be worked out by considering that there are 100cm's in 1 meter, so you need to divide 10.5 by 100. You then move the decimal place 2 places to the right (because of the two zeros), which leads you to your answer of 0.105.

A pen measuring 10.5 cm long is equivalent to 0.105 meters.

Further Explanation

Length  

  • Length refers to a measurement of distance. It is the distance between two points.

S.I units  

  • The S.I. units of lengths. However, length can be measured using other units such as millimeters, centimeters, kilometers, etc.

Metric system unit of measurement  

  • It is a system of measurement that uses units such as gram, liter, and meter to measure mass, capacity and length respectively.
  • This system measures length based on meters, such that all the other units of lengths are based on an inter-conversion of units in meters.

Conversion of units measuring length

1 centimeter = 10 millimeters  

1 meter = 100 centimeters  

1 meter = 1000 millimeters  

1 kilometer = 1000 meters  

1 kilometer = 1000000 centimeters, and so forth.

Conversion of larger units to small units  

  • To convert larger units of measurement to smaller units we multiply we multiply the number of larger units by the appropriate conversion factor.

Conversion of smaller units to larger units

  • When converting smaller units of measurements to larger units of measurements we divide the number of smaller units by the appropriate factor.
  • For example, to convert cm to m, we divide the number of cm by 100.

In the question;

The pen measures 10.5 cm  

But, 1 meter = 100 centimeters

Thus;

10.5 cm will be equivalent to;

= 10.5 cm/100 cm  

= 0.105 meters

Therefore, a pen measuring 10.5 cm in length is equivalent to 0.105 meters.

Keywords: Length, meters, centimeters, metric unit system  

Learn more about:  

Level: Middle school  

Subject: Mathematics  

Topic: Metric system of measurements  

Sub-topic: Units of measuring length  
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