All categories

3 years ago

Will mark brainliest!

1 answer:

4 0

People use data tables and graphs in many financial careers and when looking at the statistics for something as simple as what they saw on the news, etc. When scientists use them, they’re both using them to see the numbers and the facts. They’re both able to use graphs and data tables to help them. They are different though because scientists use them for science related things like how much a tree grew in a year while everyday people use them to see the average amount of drop outs per year or something along those lines.

Data tables and graphs are very vital to a scientists job. They help them easily collect and organize information to where anyone can read it. It may not be absolutely necessary, but it’s something every scientist uses.

Any scientist doing any sort of research would use them. Whether they’re a biologist, geologist or whatever, they all use graphs and data tables to help them organize their research.

You might be interested in

Write a complete set of quantum numbers for the fifth electron added to a hydrogen ion (i.e., the fifth electron in any electron

ankoles [38]

N=
l=
m(l)=
m(s)=

start with H^+ (no electrons) , then adding 5 electrons will be 1s2 2s2 2p1
so for the 5th electron
n = 2
l = 1
ml = -1
ms = 1/2

5 0

3 years ago

Read 2 more answers

Write a ground state electron configuration for each neutral atom

Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0

2 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

3 0

3 years ago

2. Which of the following is an example of velocity?

Brrunno [24]

25*5= 225 and 7hs hsueuxn udneicj

8 0

3 years ago

Read 2 more answers

At a certain temperature and pressure, 0.20 mol of carbon dioxide has a volume of 3.1 L. A 3.1-L sample of hydrogen at the same

Arada [10]

A 3.1 L sample of hydrogen d. contains the same number of molecules

as 3.1 L of carbon dioxide at the same temperature and pressure.

This is the fundamental principle of Avogadro’s hypothesis: equal volume of gases at the same temperature and pressure contain the same number of molecules.

The sample of carbon dioxide has a greater mass, a greater number of atoms, and a greater density, than the sample of hydrogen.

7 0

3 years ago