All categories

3 years ago

PHYSICAL SCIENCE!!!!HURRY

Chemistry

1 answer:

emmainna [20.7K]3 years ago

5 0

Option 1 is a suitable graph for a “constant positive velocity”.

Option: 1

Explanation:

Consider a car is moving with the constant velocity, the position and time graph with “constant positive velocity” of the car is represented by the graph which is in option 1. In this graph (option: 1) has the straight line with the slope. The slope in this “position - time” graph represents the “speed of the car”, the velocity is positive hence the graph is drawn positive direction. Here, the “slope of the car” less so the car moves with less speed and constant positive velocity.

You might be interested in

Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha

elixir [45]

Answer:

0.269 g

Explanation:

$Po_{210} ^{84}$ ⟶ $Pb_{206} ^{82}$ + $+ He_{4} ^{2}$

Data:

Half-life of $Po_{210} ^{84}$ (T(1/2)) = 138.4 days

Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol

Time = 338.8 days

Isotopic masses

$Po_{210} ^{84}$ = 209.98 g/mol

$Pb_{206} ^{82}$ = 205.97 g/mol

Concepts

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.

The radioactive decay law is

N=Noe^(-λt)

Where:

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days

λ= radioactive decay constant

The radioactive constant is related to the half-life by the next equation

λ= $\frac{ln 2}{t(1/2)}$

λ= $\frac{ln2}{138.4 days}$ =0,005008 days^(-1)

No (Atoms of $Po_{210} ^{84}$ in t=0)

To get No we need to calculate the number of atoms of $Po_{210} ^{84}$ in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of $Po_{210} ^{84}$ in the initial sample.

This is:

$\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol}$ = 0,001599 mol of PoCl4

This is the number of mol of $Po_{210} ^{84}$ in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23

0,001599 mol of $Po_{210} ^{84}$ * 6.023*10^23 atoms/ mol of $Po_{210} ^{84}$ = 9.632 *10^20 atoms of $Po_{210} ^{84}$

Atoms after 338.8 days

We use the radioactive decay law to get this value

N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

This is the number of atoms of $Po_{210} ^{84}$ in the sample after 338.8 days has passed

The number of atoms $Po_{210} ^{84}$ transformed is equal to the number of atoms of $Pb_{206} ^{82}$ produced.

The number of atoms of $Po_{210} ^{84}$ transformed is No - N

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

So, 7.866*10^20 is the number of atoms of $Pb_{206} ^{82}$ produced

We can get the mass with the Avogadro’s number

(7.866*10^20 atoms of $Pb_{206} ^{82}$ ) / ( 6.023*10^23 atoms of $Pb_{206} ^{82}$ / mol of $Pb_{206} ^{82}$ = 0.001306 moles of $Pb_{206} ^{82}$

This number of moles have a mass of:

(0,001306 moles of $Pb_{206} ^{82}$ )* (205.97 g of $Pb_{206} ^{82}$ /mol of $Pb_{206} ^{82}$ ) = 0.269 g

3 0

3 years ago

Zinc metal is added to a flask containing aqueous hydrochloric acid. The flask contains 0.400 mole of HCl. How much hydrogen gas

dusya [7]

Answer:

0.4 g of hydrogen gas would be produced.

1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.

2. HCl is the limiting reactant.

3. 2.61 g of zinc is in excess

Explanation:

From the balanced equation of reaction:

$Zn(s)+ 2HCl(aq) --> H_2(g) + ZnCl_2(aq)$

1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.

15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.

0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.

But only 0.400 mole of HCl is present. Hence, HCl is the limiting reagent. For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.

0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.

0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.

Now, since HCl is the limiting reagent;

2 moles of HCl is required to produce 1 mole of H2 according to the equation.

0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2

0.2 mole H2 = 2 x 0.2 = 0.4 g H2

Hence, 0.4 g of hydrogen gas would be produced.

6 0

3 years ago

What would be the formula for diphosphorus pentoxide ?

aleksley [76]

PbCr04 + P4O10

Hope this helps!

3 0

3 years ago

Read 2 more answers

Please only answer if you know the actual thing, don't search it up! Are these correct? :P

makkiz [27]

Hey buddy I am here to help!

1. C

2. A

3. A & B

4. C

5. C

6. A

7. A

8. A & C

Hope it helps!

Plz mark brainlist!

4 0

3 years ago

Read 2 more answers

Which of the following statements is TRUE? Question 5 options: The emission spectrum of a particular element is always the same

amm1812

Answer:

All of the above are true

Explanation:

a) The emission spectrum of a particular element is always the same and can be used to identify the element: It's true since the emission spectrum for each element is unique. It has the same bright lines at the same wavelength. This feature is used to identify elements. For example, the study of the emission spectra of light arriving from stars allow us to identify the elements presents in the star because the light contains the emission spectra of those elements.

b)The uncertainty principle states that we can never know both the exact location and speed of an electron: It is true since the velocity of an electron is related to its wave nature, while its position is related to its particle nature and we cannot simultaneously measure electron's position and velocity with precision.

c) An orbital is the volume in which we are most likely to find an electron: An orbital is a probability distribution map that is used to decribe the likely position of an electron in an atom.

5 0

4 years ago