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3 years ago

When a sample of ca(s) loses 1 mole of electrons in a reaction with a sample of o2(g) the oxygen?

2 answers:

6 0

The reaction involved in present case is:

Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)

In terms of oxidation and reduction, the reaction can be shown at

Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)

From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.

From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.

Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.

Mandarinka [93]3 years ago

3 0

Gains 1 mole of electrons

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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]

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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

If a hydrate of the formula

Lelu [443]

Answer:

It will turn red

Explanation:

Bases will turn the litmus paper (impregnated with an acid-base indicator) blue.

Acids will turn turn the litmus paper (impregnated with an acid-base indicator) red.

Since the produced HCl is a strong acid, the litmus paper will turn red, when touching the HCl.

The red shows the presence of an acid, in this case HCl.

8 0

3 years ago

From what are chemical sedimentary rocks formed?

masha68 [24]

Chemicals dissolved in water. Calcite is a good example, if I'm not mistaken.

6 0

3 years ago

Read 2 more answers

For the reaction Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)

mojhsa [17]

Answer : The value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = 151.2 kJ = 151200 J

$\Delta S^o$ = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

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

$\Delta G^o=(151200J)-(328.0K\times 169.4J/K)$

$\Delta G^o=95636.8J=95.6kJ$

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln k$

where,

$\Delta G^o$ = standard Gibbs free energy = 95636.8 J

R = gas constant = 8.314 J/K.mol

T = temperature = 328.0 K

K = equilibrium constant = ?

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

$95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k$

$k=1.70\times 10^{15}$

Therefore, the value of equilibrium constant for this reaction at 328.0 K is $1.70\times 10^{15}$

3 0

3 years ago

Consider the following equilibrium:H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-1(aq).What is the correct equilibrium expression?

Vanyuwa [196]

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:<span>

<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>

4 0

3 years ago