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denpristay [2]
2 years ago
11

Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,

NaHCO3 commonly known as baking soda. The students are using toothpicks to represent bonds and jelly beans to represent atoms. Red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms, and blue jelly beans represent sodium atoms, white jellybeans represent hydrogen atoms, like jellybeans our cruise and carbon atoms, and we jellybeans that is an oxygen atoms.
Identify the number of each jelly bean needed to make both models.
Chemistry
1 answer:
vfiekz [6]2 years ago
8 0

Answer:

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

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anyanavicka [17]

Weather can be predicted only as probable, not definite because the daily weather conditions depends on winds and storms.

It is common observation that weather forecasts are often given as a probability and never in definite terms.

This is because, the weather condition at anytime depends on the temperature of the earth's atmosphere which causes air masses to move leading to wind.

The temperature of the earth's atmosphere changes frequently hence weather conditions also change frequently.

Learn more: brainly.com/question/21209813

8 0
2 years ago
What is the molarity of a solution that contains 2.35 g of nh3 in 0.0500 l of solution?
tankabanditka [31]
The molarity is the number of moles in 1 L of the solution. 
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol 
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
3 0
2 years ago
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Choose the correct description for each of the
Aliun [14]

Answer:

Density: Physical Property

Flammability: Chemical Property

Solubility In Water: Physical Property

Reactivity With Water: Chemical Property

Melting Pot: Physical Property

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Explanation:

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7 0
2 years ago
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Consider isotopes ions protons and electrons. how many of these did dalton not discuss in his atomic theory?
MAVERICK [17]

Answer: He did not discuss about any of these.

Explanation: Dalton proposed some of the postulates for his atomic theory. They are:

1) Matter is made up of atoms which are not divisible.

2) Atoms of different elements combine in a fixed ratio to form compounds.

3) The atomic properties of given element are same including mass. This states that all the atoms of an element have same mass but the atoms of different elements have different masses.

4) No atoms are either created or destroyed during a chemical reaction.

5) Atoms of an element are identical in mass, size and other chemical and physical properties.

As it is visible from the postulates, he only discussed only about the atoms but not subatomic particles or isotopes.

8 0
2 years ago
You prepare a standard by weighing 10.751 mg of compound X into a 100 mL volumetric flask and making to volume. You further dilu
ArbitrLikvidat [17]

Answer:

0.12693 mg/L

Explanation:

First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:

  • 10.751 mg / 100 mL = 0.10751 mg/mL

Then we <u>calculate the concentration of compound X in the standard after dilution</u>:

  • 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L

Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:

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Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:

  • 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
4 0
1 year ago
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