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3 years ago

What does a straight line represent on a distance vs time graph?

Chemistry

2 answers:

bija089 [108]3 years ago

4 0

Answer:

It represents the relationship between the distance travelled and time spent as constant or at a stop.

Explanation:

This is because as the line progresses, the time increases the distance should change as well my moving up or down, in a linear equation this relationship is constand but if the line is horizontal though, the distance is not changing despite time passing which means it has come to a stop.

stellarik [79]3 years ago

3 0

A straight line on a distance va time graph represents constant speed

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Write balanced nuclear equations for the formation of five elements whose atomic number is between helium (2) and iron (26).

stiv31 [10]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

 the equation for helium could be: D + T -> He + n. D is deuterium (H-2), T is tritium (H-3), and n symbolizes a neutron.

3 0

3 years ago

1. If you have 8.1 g CO₂ and 2.3 g H₂, which one is the limiting reactant. Show your work. 2. Calculate the theoretical yield of

ZanzabumX [31]

Answer:

2.95 g of CH₄

Explanation:

To start this, we determine the equation:

4H₂ + CO₂ → CH₄ + 2H₂O

4 moles of hydrogen react to 1 mol of carbon dioxide in order to produce 1 mol of methane and 2 moles of water.

To determine the limiting reactant, we need to know the moles of each reactant.

8.1 g . 1 mol/ 44g = 0.184 moles of carbon dioxide

2.3 g . 1mol / 2g = 1.15 moles of hydrogen

4 moles of hydrogen react to 1 mol of CO₂

Then, 1.15 moles may react to (1.15 . 1) /4 = 0.2875 moles

We only have 0.184 moles of CO₂, so this is the limiting reactant. Not enough CO₂ to complete the 0.2875 moles that are needed.

Ratio is 1:1. 1 mol of CO₂ produces 1 mol of methane

Then, 0.184 moles of CO₂ will produce 0.184 moles of CH₄

We convert moles to mass: 0.184 mol . 16 g /mol = 2.95 g

7 0

3 years ago

Devices that allow you to do work in an easier way.

IrinaVladis [17]

Answer:

The most notable of these are known as the "six simple machines": the wheel and axle, the lever, the inclined plane, the pulley, the screw, and the wedge, although the latter three are actually just extensions or combinations of the first three.

6 0

3 years ago

Read 2 more answers

The fertilizer ammonium sulfate, (NH4)2SO4, is prepared by the reaction between ammonia (NH3) and sulfuric acid:

kumpel [21]

The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is 6.18 * 10⁴ Kg of ammonia.

<h3>What mass in kilograms of ammonia are required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄?</h3>

The mass of ammonia required to produce 2.40 × 10⁵ kg of (NH₄)₂SO₄ is determined from the mole ratio of the reaction.

The mole ratio of the reaction is obtained from the balanced equation of the reaction given below:

2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)

Mole ratio of NH₃ and (NH₄)₂SO₄ is 2: 1

Mass of 2 moles of ammonia = 2 * 17 = 34 g

Mass of 1 mole of (NH₄)₂SO₄ = 132 g

Mass of ammonia required = 34/132 * 2.40 × 10⁵ kg

Mass of ammonia required = 6.18 * 10⁴ Kg of ammonia.

In conclusion, the mole ratio is used to determine the mass of ammonia required.

Learn more about mole ratio at: brainly.com/question/19099163

#SPJ1

4 0

2 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago