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3 years ago

The temperature of a boiling liquid can rise above its boiling point. a. true b. false

2 answers:

5 0

The temperature of a liquid can exceed its boiling point. An example is water. Although at ordinary pressure of 1 atm, the boiling point is 100 degrees, water can still exist in higher temperatures but this time in another state. Superheated steam is the term used for water whose temperature has higher than the boiling point

jekas [21]3 years ago

5 0

I don't know if you got your answer yet but its false. I just took the test and it was false.

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What represents a community?<br> pond<br> B<br> a snail<br> snails<br> D snails, guppies and plants

OLga [1]

D snails, guppies and plants. The community should be diverse and have lots of biodiversity.

7 0

3 years ago

Which of the following pairs of reactants will react together to produce water (H2O) as one of the products?

Rainbow [258]

1) <span>NaNO3 and H2O - no reaction , it is dissolution
2) no hydrogen to make water
3) </span><span>Fe(OH)3 (base) and H2SO4(acid))
base +acid ----> salt +water
4) </span><span>Li2O and Ba(OH)2
basic oxide and base ----> no reaction

so Answer number 3)
</span> 2Fe(OH)3 +3 H2SO4 ------> Fe2(SO4)3 + 6H2O<span>

</span>

7 0

3 years ago

Read 2 more answers

You make an experiment to determine the percent mass of carbon in sugar. You find out that there is 46.63g of carbon in 75.00g o

Aleksandr [31]

Taking into account the definition of percentage composition, the percent composition of carbon in this sample is 62.17%.

<h3>Percentage composition</h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

Then, the percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$percentageby mass= \frac{mass of solute}{mass of solution}x100$

In this case, you know:

mass of solute= 46.63 g
mass of solution= 75 g

Replacing:

$percentageby mass= \frac{46.63 g}{75 g}x100$

Solving:

<u><em>percentage by mass= 62.17 %</em></u>

Finally, the percent composition of carbon in this sample is 62.17%.

Learn more about percent composition:

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<u>brainly.com/question/9779410?referrer=searchResults</u>

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3 0

3 years ago

Think about the lab procedure you just read. Label each factor below with V for “variable” or C for “constant”.

luda_lava [24]

<h2>Answer:</h2><h3>The temperature of the gas: V</h3>

The temperature of gas is a variable quantity. It can be changed by changing energy or pressure of gas.

<h3>The amount of gas in the tube (in terms of mass and moles): C</h3>

It is a constant entity. As mass of gas once taken can not be changed by changing temperature, pressure etc.

<h3>The radius of the tube: C</h3>

The radius of tube cannot change at any rate.

<h3>The temperature of the gas (changed by the water surrounding it): V</h3>

It can be changed by changing the temperature of water surrounding it.

It can never be changed.

<h3>The pressure of the gas: V</h3>

It can be changed by simply changing temperature and volume of gas.

8 0

4 years ago

Read 2 more answers

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sam

elena-14-01-66 [18.8K]

Answer:

The empirical formula is C4H5ClO2

Explanation:

Step 1: Data given

Mass of the sample = 40.10 grams

Mass of CO2 produced = 58.57 grams

Mass of H2O produced = 14.98 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C= 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

In experiment 2, mass = 75.00 grams and 22.06 grams is Cl

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 58.57 grams / 44.01 g/mol

Moles CO2 = 1.33 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol CO2

For 1.33 moles CO2 we have 1.33 moles C

Step 4: Calculate mass C

Mass C = 1.33 grams * 12.01 g/mol

Mass C = 15.97 grams

Step 5: Calculate moles H2O

Moles H2O= 14.98 grams /18.02 g/mol

Moles H2O = 0.831 moles

Step 6: Calculate moles H

For 1mol H2O we have 2 moles H

For 0.831 moles H2O we have 2*0.831 = 1.662 moles H

Step7: Calculate mass H

Mass H = 1.662 moles * 1.01 g/mol

Mass H = 1.68 grams

Step 8: Calculate mass %

%C = (15.97 grams / 40.10) * 100 %

%C = 39.8 %

%H = (1.68 / 40.10 ) *100%

%H = 4.2 %

%Cl = (22.06 / 75.00 ) * 100%

%Cl = 29.4 %

%O = 100 % - 39.8% - 4.2 % - 29.4 %

%O = 26.6 %

Step 9: Calculate moles in compound

We assume the compound has a mass of 100 grams

Mass C = 39.8 grams

MAss H = 4.2 grams

MAss Cl = 29.4 grams

Mass O = 26.6 grams

Moles C = 39.8 grams / 12.01 g/mol

Moles C = 3.314 moles

Moles H = 4.2 moles / 1.01 g/mol

Moles H = 4.158 moles

Moles Cl =29.4 grams / 35.45 g/mol

Moles Cl = 0.829 moles

Moles O = 26.6 grams / 16.0 g/mol

Moles O = 1.663 moles

Step 10: calculate the mol ratio

We divide by the smallest amount of moles

C: 3.314 moles / 0.829 moles = 4

H: 4.158 moles / 0.829 moles = 5

Cl: 0.829 moles /0.829 moles = 1

O: 1.663 moles / 0.829 moles = 2

This means For each Cl atom we have 4 C atoms, 5 H atoms and 2 O atoms

The empirical formula is C4H5ClO2

5 0

3 years ago