Which methods could be used to dilute a solution of sodium chloride (NaCl)?

A 28.5 gram piece of iron is added to a graduated cylinder containing 45.5 mL of water. The water in the cylinder rises to the 4

sweet-ann [11.9K]

Answer:

7.92 .g . mL^-1

Explanation:

Data provided in the question

Water liters = 45.5 mL

Pace of iron = 28.5 gram

Rise in the water in the cylinder = 49.1 mark

So by considering the above information

As we know that

$Density = \frac{Mass}{Volume}$

where,

Mass = 28.5 gram

Volume = 49.1 - 45.5

= 3.6 mL

So, the density is

$Density = \frac{28.5}{3.6}$

= 7.92 .g . mL^-1

Basically we use the density formula by dividing the mass with the volume

4 0

3 years ago

Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

klio [65]

Answer:

10%

Explanation:

Given data:

Actual yield = 0.13 moles

Percent yield = ?

Solution:

Balanced chemical equation.

3H₂ + N₂ → 2NH₃

First of all we will calculate the mass of given moles of ammonia.

Number of moles = mass / molar mass

Mass = 0.13 × 17 g/mol

Mass = 2.21 g

2.21 g is experimental yield of ammonia.

Hydrogen is limiting reactant because only two moles of hydrogen present.

we will compare the moles of hydrogen and ammonia from balance chemical equation.

H₂ : NH₃

3 : 2

2 : 2/3×2 = 1.33 moles

Mass of ammonia

Mass = number of moles × molar mass

Mass = 1.33 mol × 17 g/mol

Mass = 22.61 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 2.21 g / 22.61 g

Percent yield =9.8% which is almost 10%

3 0

3 years ago

Which describes any substance that shatters or breaks easily? malleable brittle volatile ductile

swat32

Answer:

b

Explanation:

ididthetest

5 0

2 years ago

Read 2 more answers

A sample of a hydrocarbon contains 20.75 g C and 4.25 g H. Its molar mass is 58.04 g/mol. What is its empirical formula C2H5 CH2

Xelga [282]

It is c2h5 if you our doing it on edunitey thats all i can say

6 0

3 years ago

Read 2 more answers

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