Calculate number of moles of 4.4g of CO2 and 5.6L of NH3 . Pls help . Its urgent

An element X exists in three forms A, B and C in the ratio 1:2:3. If C has 10 protonsand the number of neutrons in A, B and C ar

Strike441 [17]

x21 +ANSWER

(i) Ne-22

(ii)1s2s22p6

(iii)21.3

An element X exists in three forms A, B and C in the ratio 1:2:3. If C has 10 protons and the number of neutrons in A, B and C are 10, 11 and 12 respectively,Give the following:(i) Representation of form C of the element X(ii) Electronic configuration of form B of the element(iii) Calculate the average atomic mass.

(i)C has 10 protons and 12 neutrons so a mass of 10 +12 =22

element 10 is Neon (Ne) so this isotope is Ne-22

(ii) they all have the sane atomic number so the same number of electrons

with an electronic structure of 1s2s22p6

(iii) A weighs 20, B weighs 21, C weighs 22

the ratio is 1:2:3

weighted average weight is therefor

(1X20 +2X21 +3X22)/6 =21.3

4 0

3 years ago

1) A car is traveling down the interstate at 37.1 m/s. The driver sees a cop and quickly slows down. If the driver slows to 29.8

WITCHER [35]

1)

The acceleration of the car is the rate of change of velocity of the car; it can be calculated as:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity of the car to change from u to v

In this problem, for this car we have:

u = 37.1 m/s

v = 29.8 m/s

t = 3 s

So, the acceleration is:

$a=\frac{29.8-37.1}{3}=-2.43 m/s^2$

2)

The work done in lifting the box is equal to the potential energy transferred to the box during the process; it is given by:

$W=Fd$

where

F is the force applied

d is the displacement of the box

Here we have:

F = 87.3 N is the force applied

d = 2.04 m is the displacement of the box

So, the work done to lift the box is:

$W=(87.3)(2.04)=178.1 J$

3)

The power is the rate of work done per unit time. It is calculated as:

$P=\frac{W}{t}$

where

W is the work done

t is the time taken to do the work

For the child in this problem, we have:

W = 1250 J is the work done by the child running up the stairs

P = 267 W is the power used

Therefore, re-arranging the equation, we find the time taken:

$t=\frac{W}{P}=\frac{1250}{267}=4.68 s$

4)

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the rabbit in this problem, we have:

m = 8.642 kg is the mass of the rabbit

KE = 125.6 is its kinetic energy

Solving the formula for v, we find the speed of the rabbit:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(125.6)}{8.642}}=5.4 m/s$

5)

The efficiency of a machine is the ratio between the energy produced in output by the machine and the work done in input. Mathematically, it is given by

$\eta = \frac{E_{out}}{W_{in}}\cdot 100$

where

$E_{out}$ is the energy in output

$W_{in}$ is the work in input

For the machine in this problem,

$W_{in}=120 J$ is the work in input

$E_{out}=93 J$ is the energy in output

Therefore, the efficiency of this machine is:

$\eta=\frac{93}{120}\cdot 100=77.5\%$

6)

During a collision, the total momentum of the system is always conserved before and after the collision. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 =(m_1+m_2)v$

where

$m_1=212 kg$ is the mass of the first car

$u_1=8.00 m/s$ is the initial velocity of the first car

$m_2=196 kg$ is the mass of the 2nd car

$u_2=6.75 m/s$ is the initial velocity of the 2nd car

$v$ is the final velocity of the two cars stuck together (after the collision, they move together)

Solving the equation for v, we find:

$v=\frac{m_1 u_1 +m_2 u_2}{m_1 +m_2}=\frac{(212)(8.00)+(196)(6.75)}{212+196}=7.40 m/s$

7)

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed of the wave

f is the frequency of the wave

$\lambda$ is the wavelength

For the wave in the string in this problem we have:

$\lambda=0.23 m$ (wavelength)

f = 12 Hz (frequency)

So, the speed of the wave is:

$v=(12)(0.23)=2.76 m/s$

8)

The relationship between frequency and wavelength for an electromagnetic wave is given by

$c=f\lambda$

where:

c is the speed of light in a vacuum

f is the frequency of the wave

$\lambda$ is the wavelength of the wave

For the blue light in this problem, we have

$f=6.2\cdot 10^{14}Hz$ (frequency)

while the speed of light is

$c=3.0\cdot 10^8 m/s$

So, the wavelength of blue light is:

$\lambda=\frac{c}{f}=\frac{3.0\cdot 10^8}{6.2\cdot 10^{14}}=4.8\cdot 10^{-7} m$

9)

The sound wave in this problem travels with uniform motion (=constant velocity), therefore we can use the following equation:

$d=vt$

where

d is the distance covered by the wave

v is the speed of the wave

t is the time elapsed

In this problem:

v = 343 m/s is the speed of the sound wave

t = 0.287 s is the time elapsed

So, the distance covered by the wave is

$d=(343)(0.287)=98.4 m$

3 0

3 years ago

Complete the paragraph about the formation of sodium phosphide.

Nadya [2.5K]

<h2>Let us complete it :</h2>

Explanation:

let us study about formation of sodium phosphide

Electrons are transferred from atoms of sodium to atoms of phosphorous

The sodium atom looses electrons and the phosphorus atoms

gains electrons .

C. This transfer makes the sodium atoms acquire positive charge and phosphorous acquire negative charge .

As a result, the sodium and phosphorus atoms strongly bond with

each other.

8 0

3 years ago

In a chemical equation, the arrow

muminat

In a chemical equation, the arrow

A. can be read as "yields" or "makes."

B. always points toward the products.

C. separates the products and reactants.

D. all of these

all of these options are right.

7 0

3 years ago

Read 2 more answers

The central Xe atom in the XeF4 molecule has ________ unbonded electron pair(s) and ________ bonded electron pair(s) in its vale

Dimas [21]

Answer:

Lewis structure in attachment.

Explanation:

Atoms of elements in and beyond the third period of the periodic table form some compounds in which more than eight electrons surround the central atom. In addition to the 3s and 3p orbitals, elements in the third period also have 3d orbitals that can be used in bonding. These orbitals enable an atom to form an <u>expanded octet</u>.

The central Xe atom in the XeF₄ molecule has <u>two</u> unbonded electron pairs and <u>four</u> bonded electron pairs in its valence shell.

8 0

3 years ago