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IRINA_888 [86]
3 years ago
7

The variable that a scientist observes to change while conducting an experiment is called the ___ variable.

Chemistry
2 answers:
QveST [7]3 years ago
5 0

The variable that a scientist observes to change while conducting an experiment is called the responding variable.

Nimfa-mama [501]3 years ago
4 0

Great Question!

The Answer Would Be "B" The "RESPONDING" Variable

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If 23.6 g of hydrogen gas reacts with 28.3 g of nitrogen gas, what is the maximum amount of product that can be produced?
Aleonysh [2.5K]

Answer:

34.3 g NH3

Explanation:

M(H2) = 2*1 = 2 g/mol

M(N2) = 2*14 = 28 g/mol

M(NH3) = 14 + 3*1 = 17 g/mol

23.6 g H2* 1 mol/2 g = 11.8 mol H2

28.3 g N2 * 1 mol/28 g = 1.01 mol N2

                                 3H2 + N2 ------> 2NH3

from reaction         3 mol    1 mol

given                   11.8 mol    1.01 mol

We can see that H2 is given in excess, N2 is limiting reactant.

                                 3H2 + N2 ------> 2NH3

from reaction                     1 mol         2 mol

given                                 1.01 mol      x

x = 2*1.01/1= 2.02 mol NH3

2.02 mol * 17g/1 mol ≈ 34.3 g NH3

8 0
3 years ago
A 25.0 mL sample of 0.100 M lactic acid (HC3H503, Ka = 1.4 E-4) is titrated with 0.100 M NaOH solution. Calculate the pH of the
aleksklad [387]
I’m a bit confused too
4 0
3 years ago
Which model failed to explain the
11Alexandr11 [23.1K]

Answer:

C). The Bohr-Rutherford model

Explanation:

The 'Bohr-Rutherford model' of the atom failed to elaborate on the attraction between some substances. It essentially targeted hydrogen atoms and failed to explain its stability across multi-electrons. The nature and processes of the chemical reactions remained unillustrated and thus, this is the key drawback of this model. Thus, <u>option C</u> is the correct answer.

5 0
3 years ago
What is the oxidation state of an individual carbon atom in caco3?
Marizza181 [45]

Answer:

The oxidation state of the carbon is +4.

Explanation:

Calcium is in group 2 of the periodic table, therefore, its oxidation state is +2.

The oxidation state of the oxygen is -2.

As the compound is neutral, the sum of the oxidation states of all atoms must be 0.

Oxidation State Ca + Oxidation State C + (Oxidation State O)×3 = 0

+2 + x + (-2)×3 = 0

2 + x - 6 = 0

x = 6 -2

x = 4

Hence, the oxidation state of the carbon is +4.

4 0
3 years ago
A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure
ycow [4]

<u>Answer:</u> The molecular formula for the compound is C_6H_6

<u>Explanation:</u>

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

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

Mass of C = 92.26 g

Mass of H = 7.74 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles

  • <u>Step 2:</u> 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 7.68 moles.

For Carbon = \frac{7.68}{7.68}=1

For Hydrogen = \frac{7.74}{7.68}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is CH

  • <u>Calculating the molar mass of the compound:</u>

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

PV=\frac{m}{M}RT

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L  (Conversion factor:  1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = 62.3637\text{ L. torr }mol^{-1}K^{-1}

T = temperature of the gas = 100^oC=(100+273)K=373K

Putting values in above equation, we get:

820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol

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 = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

n=\frac{78.10g/mol}{13g/mol}=6

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

C_{(1\times 6)}H_{(1\times 6)}=C_6H_6

Hence, the molecular formula for the compound is C_6H_6

8 0
3 years ago
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