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natima [27]
3 years ago
8

1. A student represented a water molecule using beads and straws. She connected two black beads and one white bead in a certain

manner using straws. Which of the following explains the type of scientific model created by the student?
A. A simulation because it helps visualize the shape of the molecule.
B. A 3-dimensional model because it helps visualize the shape of the molecule.
C. A simulation because it helps understand the movement of atoms in the molecule.
D. A 3-dimensional model because it helps understand the movement of atoms in the molecule.
Chemistry
2 answers:
Rina8888 [55]3 years ago
7 0

Hello! Your answer is (B. A 3-dimensional model because it helps visualize the shape of the molecule.) This is because the model they are creating is not a simulation. Also, it can help us visualize the shape of the molecule due to the manner in which it was formed.

mars1129 [50]3 years ago
3 0

Answer: Option (b) is the correct answer.

Explanation:

When the student is using beads then he is representing atoms and the straws represent the bond between the atoms.

Therefore, when a model is represented by the given manner then it helps in identifying the geometry or shape of the object. This is because it shows how atoms are arranged in space along with their angle.

Therefore, we can conclude that out of the given options the type of scientific model created by the student is a 3-dimensional model because it helps visualize the shape of the molecule.

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5 0
3 years ago
Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo
hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

q=mc\Delta T         .........(1)

where,

q = heat absorbed

m = mass of substance

c = specific heat capacity of substance

\Delta T = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

q=m\times \Delta H      ........(2)

where,

q = heat absorbed

m = mass of substance

\Delta H = enthalpy of the reaction

The processes involved in the given problem are:

1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K

Putting values in equation 1, we get:

q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J

  • <u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J

Total heat required = [q_1+q_2]

Total heat required = [4153.8J+32467.7J]=36621.5J

Hence, the total heat required is 36621.5 J

  • <u>For b:</u>

The processes involved in the given problem are:  

1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K

Putting values in equation 1, we get:

q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J

  • <u>For process 2:</u>

We are given:

m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g

Putting values in equation 2, we get:

q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J

  • <u>For process 3:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K

Putting values in equation 1, we get:

q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J

  • <u>For process 4:</u>

We are given:

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J

Total heat required = [q_1+q_2+q_3+q_4]

Total heat required = [448.14+4583.5+21445.2+32467.7]J=58944.5J

Hence, the total heat required is 58944.5 J

8 0
3 years ago
Which of the following would be a clue that a rock is sedimentary?​
Masteriza [31]
I do not see any possible answers that you have posted. But a good idea would be that it might have fossils, water currents, mud cracks, usually form in shallow seas. Hope this helps!
8 0
3 years ago
Read 2 more answers
Compare and contrast steak and juice​
blsea [12.9K]
Steak is a meat and juice is a liquid
5 0
3 years ago
2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

732.767 g CO₂ ≈ 733 g CO₂

8 0
3 years ago
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