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2 years ago

Acceleration always refers to a(n)

Chemistry

2 answers:

Harman [31]2 years ago

8 0

B, increase in speed

konstantin123 [22]2 years ago

6 0

Answer:

increase in speed

Step-by-step explanaton:

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A newly discovered element, Z, has two naturally occurring isotopes. 90.3 percent of the sample is an isotope with a mass of 267

blsea [12.9K]

atomic mass=percentage of isotope a * mass of isotope a + percentage of isotope b * mass of isotope b+...+percentage of isotope n * mass of isotope n.

Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%

mass of isotope₂=270.9 u
percentage of isotope₂=9.7%

Therefore:

atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u

Answer: the mass atomic of this element would be 268.1 u

7 0

3 years ago

Why is it important to make observations about the reactants?

Free_Kalibri [48]

Answer:

Reason Down below

Explanation:

It is important because when you make observation you get a clue sometimes and it reactants i feel like it also takes places with observation. :)

7 0

3 years ago

A metal ion Mⁿ⁺ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 2.961 x 10¹⁶ Hz

Whitepunk [10]

Answer:

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

Explanation:

First of all

v=f*λ

In our case v=c

c=f*λ

λ=c/f

where:

c is the speed of light

f is the frequency

$\lambda=\frac{3*10^8}{2.961*10^{16}}\\ \lambda=1.01317*10^{-8} m$

Using Rydberg's Formula:

$\frac{1}{\lambda}=R*z^2*(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where:

R is Rydberg constant= $1.097*10^7$

z is atomic Number

For highest Energy:

n_1=1

n_2=∞

$\frac{1}{1.01317*10^{-8}}=1.097*10^{7}*z^2*(\frac{1}{1^2}-\frac{1}{\inf})\\z^2=8.99\\z=2.99$

z≅3

Atomic number is 3, So ion is Lithium ion ( $Li^+$ )

3 0

3 years ago

What are the three heredity rules ?

Nataly [62]

The Law of Dominance. ...
The Law of Segregation. ...
The Law of Independent Assortment.

7 0

3 years ago

Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of

DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization = 35.2 kJ/mol

Changing to J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm) = 4,233.822468 x-0.0005466866

ln (P2/ 1 atm)= -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm x760 = 75.09 torr

7 0

2 years ago