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liraira [26]
3 years ago
8

In an investigation with density, Marcia’s teacher measures the mass and volume of 10 different samples of a substance. The samp

les are numbered 1 through 10. Then, the teacher makes the graph that is shown below. What is the independent variable in this investigation? mass volume sample number substance density

Chemistry
2 answers:
valina [46]3 years ago
3 0

Answer:

Mass  

Step-by-step explanation:

Usually, you plot the independent variable along the horizontal (x) axis and the dependent variable along the vertical (y) axis.

Marcia's teacher plotted the mass of the sample along the x-axis and volume along the y-axis.

The mass is the independent variable, because that is <em>what the teacher varied</em>.

The volume is the <em>dependent variable</em>, because it <em>depends</em> on the mass.

Sample number is <em>wrong</em>, because it is not a variable.

Substance is <em>wrong</em>, because all samples consist of the same substance.

Density is <em>wrong</em>, because it is constant. It is the slope of the graph.

emmainna [20.7K]3 years ago
3 0

Answer:mass

Explanation:

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How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.
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<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

   N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

                                  = 2.3 moles × 2

                                  = 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

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Therefore;

Mass = 4.6 moles × 17.031 g/mol

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3 0
3 years ago
How many kJ of heat are required to melt 50.0 g of ice at its melting pt?
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5 0
3 years ago
a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what i
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Answer:

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Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

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H = 8.1636%

O = 43.1981%

Divide by their molar mass

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Divide by the smallest

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H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

8 0
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