1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
weqwewe [10]
3 years ago
5

What mass of oxygen is contained in a 5.8 g sample of NaHCO3?

Chemistry
1 answer:
jarptica [38.1K]3 years ago
4 0
X:5.8g=16:(23+1+12+3*16)
x:5.=16:84 
x:=5.8* 16/84 
this is approximately 1.1
You might be interested in
Which of the following is an example of a solution? A. An iron alloy B. Iron metal C. An iron ore D. Iron rust
Ganezh [65]
The one that is an example of a solution is : A. an iron alloy
iron alloy consist of minor component that is distributed within the major component, which is the basic requirement for a solution
4 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
slega [8]

Answer:

The catalyzed reaction will take time of 5.11\times 10^{-8} years.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate of reaction with catalyst

K_1 = rate of reaction without catalyst

Ea_2 = activation energy with catalyst  = 59.0 kJ/mol = 59000 J/mol

Ea_1 = activation energy without catalyst  = 184 kJ/mol = 184000 J/mol

R = gas constant

T = temperature = 600K

Now put all the given values in this formula, we get:

\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}

The reaction enhances by 7.632\times 10^{10}  when catalyst is present.

Time taken by reaction without catalyzed = 3900 years

Time taken by reaction with catalyzed = x

x=\frac{3900 year}{7.632\times 10^{10}}=5.11\times 10^{-8} years

The catalyzed reaction will take time of 5.11\times 10^{-8} years.

8 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
How many liters of methane gas (CH4) need to be combusted to produce 12.4 liters of water vapor, if all measurements are taken a
Ivan
The balanced chemical reaction is written as:

<span>CH4 (g) + 2 O2 (g) ----> CO2 (g) + 2 H2O (g)
</span>
We are given the amount of water to be produced from the reaction. This amount will be used for the calculations. Calculations are as follows:

12.4 L H2O ( 1 mol / 22.4 L ) ( 1 mol CH4 / 2 mol H2O ) ( 22.4 L / 1 mol ) = 6.2 L CH4
7 0
3 years ago
Which substance may lower air temperatures after a volcanic eruption? lava O sulfur dioxide O carbon dioxide O water vapor​
IrinaVladis [17]
I think it’s sulfur dioxide
8 0
3 years ago
Read 2 more answers
Other questions:
  • How many grams of H3PO4 are produced when 10.0 moles of water react with an excess of P4O10?
    6·1 answer
  • What are weak bonds that allow flexibility in enzymes
    8·1 answer
  • A photon with a wavelength of less than 50.4 nm can ionize a helium atom.
    13·1 answer
  • a mixture is made by combining 1.64 lb of salt and 4.66 lb of water. what is the percentage of salt (by mass) in this mixture?
    7·2 answers
  • THIS IS ON MY UNIT 3 STUDY GUIDE!!!!!!!!!<br><br> What is a covalent compound? Write an example.
    13·1 answer
  • What colors are absorbed Best Buy the plant gizmo
    7·1 answer
  • Please help fast what is this molecule called
    14·1 answer
  • What type of energy is best described as the energy required to break the bonds present in the reactants?
    5·1 answer
  • How many atoms are in 15.1g of potassium?
    13·1 answer
  • which ionic bond is stronger: the ionic bond between calcium ions and chloride ions in crystalline calcium chloride , or the ion
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!