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3 years ago

On a position-versus-time graph, where is time usually shown?

Chemistry

2 answers:

Ivahew [28]3 years ago

6 0

Answer:

On a position-versus-time graph, along the x-axis (option b) is time usually shown.

Explanation:

One way to describe and study the movements is by means of graphs that represent distance-time (distance as a function of time), speed-time (speed as a function of time) and acceleration-time (acceleration as a function of time).

Then, as the displacements are made while time passes, the description of the movement is facilitated by making a graph of position versus time. In the vertical axis the positions that the body occupies are usually represented and in the horizontal axis the time. Similarly, velocity or acceleration can be represented versus time, where the horizontal axis is time and the vertical axis is velocity or acceleration.

Taking into account that the horizontal axis belongs to the x axis and the vertical axis to the y axis:

On a position-versus-time graph, along the x-axis (option b) is time usually shown.

Dennis_Churaev [7]3 years ago

5 0

time is usually on the x axis while position is on the y axis>

hope this helped ^_^

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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00$

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

$n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00 \approx 5$

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

8 0

4 years ago

Which part of the mantle is still solid but flows like a heavy liquid

topjm [15]

The Earth is composed of four different layers. Many geologists believe that as the Earth cooled the heavier, denser materials sank to the center and the lighter materials rose to the top. Because of this, the crust is made of the lightest materials (rock- basalts and granites) and the core consists of heavy metals (nickel and iron).

The crust is the layer that you live on, and it is the most widely studied and understood. The mantle is much hotter and has the ability to flow. The Outer and Inner Cores are hotter still with pressures so great that you would be squeezed into a ball smaller than a marble if you were able to go to the center of the Earth!!!!!!

5 0

3 years ago

A sample of a compound containing boron (B) and hydrogen (H) contains 5.443 g of B and 1.522 g of H. The molar mass of the compo

Svetach [21]

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

first you divide the mass of BORON by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass.

So, 5.443/11 is about 0.5. Then, the RFM of H is 1, so the number of mole is 1.522/1=1.522.

Next, you get the number of moles in order is; 0.5 and 1.522. Now we need to look at the ratio between these numbers. 0.5 is smaller so we use it as the ratio of 1. next use 1.522/0.5 is 3.044 which has a greatest common factor of 3. so the empirical formula is BH3.

Now we are going to solve the molecular formula.

the molar mass ofthis compound is 30g, so we're going to find the RFM of the empirical formulsof BH3 first.

11+3=14.

now we see how many times 14 goes into 30. 30/14=2.14 which is about 2.

So now we need to times the subscript of the empirical formula by two.

thus, the molecular formula is B2H6.

To solve this kind of questions, there are many steps:Know what you are calculating about, it's about the molecular formula, so you need to find out the number of moles of each elements. then use the molar mass of the whole compound to calculate the molecular formula.

1) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4) then use the molarmass of the whole compound to find the molecular formula. molar mass of the whole compound/RFM(molar mass) of the empirical formula of elements= the number you need to multiply by the subscript of the empirical formula to get the molecular formula.

please tell me if i got anything wrong;)

4 0

3 years ago

Which two types of information are found in an element's box in the periodic

asambeis [7]

Answer:

The answer is B and D.

Explanation:

prove me wrong

4 0

3 years ago

Read 2 more answers

Am I correct ! Please tell me. If not then pls correct ! Thanks

scZoUnD [109]

Cu + 2H2SO4 ⟶ CuSO4 + SO2 + 2H20

In left hand side of the equation.

Cu = 1 atom

H = 4 atoms

S = 2 atoms

O = 8 atoms

In right hand side of the equation.

Cu = 1 atom

S = 2 atoms

0 = 8 atoms

H = 4 atoms

• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.

• Equation is balanced and correct.

*See the attachment .

3 0

2 years ago