What is the name of Ce element in periodic table?

How do you get bic grip permanent off of your hands?

Montano1993 [528]

You could try rubbing coconut oil on it with a cloth it should come off

4 0

3 years ago

Read 2 more answers

How to find percentage abundanace

Ghella [55]

Answer and Explanation:

How to Calculate the Percent Abundance of an Isotope

Step 1: Find the Average Atomic Mass. Identify the atomic mass of the element from your isotopic abundance problem on the periodic table. ...

Step 2: Set Up the Relative Abundance Problem. ...

Step 3: Solve for x to Get the Relative Abundance of the Unknown Isotope. ...

Step 4: Find percent abundance.

5 0

4 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

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

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

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

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

4 years ago

Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze

Rashid [163]

Answer:

a. Δ $H^0_{rxn} = -108.0\frac{kJ}{mol}$

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate Δ $H^0_{rxn}$ , for the gas-phase reaction ) using the Hess's Law.

Δ $H^0_{rxn}$ = $E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}$

Given from the question, the table below shows the corresponding Δ $H^0_{t}(kJ/mol)$ for each compound.

Compound $H^0_{t}(kJ/mol)$

Liquid EO -77.4

$CH_4_(g_)$ -74.9

$CO_(g_)$ -110.5

If we incorporate our data into the above previous equation; we have:

Δ $H^0_{rxn}$ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

= $-108.0 \frac{kJ}{mol}$

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

$T_{initial}$ = 93.0°C &

the enthalpy of vaporization (Δ $H^0_{vap}$ ) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as = $\frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}$

Let's not forget as well, that Δ $H^0_{vap}$ = $\frac{q}{mass}$

If we substitute Δ $H^0_{vap}$ for $\frac{q}{mass}$ in the above equation, we have;

specific heat capacity (c) = $\frac{deltaH^0_{vap}}{T_{final}-T_{initial}}$

Making ( $T_{final}- T_{initial}$ ) the subject of the formula; we have:

$T_{final}- T_{initial}$ = $\frac{delat H^0_{vap}}{specificheat capacity}$

$(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}$

$T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C$

= 227.76°C +93.0°C

= 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C

7 0

4 years ago

Tim wants to store 0.122 moles of helium gas at 203 kPa and 401 K. Which is the volume of the helium gas needed for storage?

algol13

The volume of Helium gas needed for storage is 2.00 L (answer C)

<u><em> calculation</em></u>

The volume of Helium is calculated using ideal gas equation

That is Pv =nRT

where;

P( pressure) = 203 KPa

V(volume)=?

n(number of moles) = 0.122 moles

R(gas constant) = 8.314 L.Kpa/mol.K

T(temperature)= 401 K

make V the subject of the formula by diving both side by P

V=nRT/p

V={[0.122 moles x 8.314 L. KPa/mol.K x 401 K] / 203 KPa} = 2.00 L

3 0

3 years ago

What is the name of Ce element in periodic table?​

What is the name of Ce element in periodic table?