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3 years ago

An unknown element X has the following isotopes: 126X (22.00%

Chemistry

1 answer:

enot [183]3 years ago

6 0

Answer:

$m_X=128.44 a.m.u$

Explanation:

Hello there!

In this case, since the average atomic mass of an element, when given the atomic mass and the percent abundance of the naturally occurring isotopes is computed as shown below for this unknown element:

$m_X=126*0.22+128*0.34+130*0.44$

Whereas the atomic mass of each isotope is multiplied by the percent abundancy; we obtain:

$m_X=128.44 a.m.u$

Best regards!

You might be interested in

in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co

kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

$\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5$

mol of element G :

$\tt mol=\dfrac{4}{16}=0.25$

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0

2 years ago

Avogradro's number is the number of particles in one gram of carbon- 12 atom true or false?

LiRa [457]

Answer:

True

Explanation:

The value of the mole is equal to the number of atoms in exactly 12 grams of pure carbon-12. 12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).

5 0

3 years ago

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

Best regards.

8 0

3 years ago

PLEASE HELP ME ASAPPPP

sattari [20]

Boyle’s law gives the relationship between pressure and volume of gases. It states that at constant temperature the pressure of gas is inversely proportional to volume of gas.
PV = k
Where P is pressure V is volume and k is constant
P1V1 = P2V2
Parameters at STP are on the left side and parameters for the second instance are on the right side of the equation
P1 - standard pressure - 1.0 atm
Substituting the values in the equation
1.0 atm x 5.00 L = P x 15.0 L
P = 0.33 atm
New pressure is 0.33 atm

5 0

3 years ago

The reaction between A2 (two large red spheres) and B2 (two small blue spheres) to produce AB is shown in the diagram. The react

drek231 [11]

Answer:

B₂

Explanation:

The limiting reactant is always a reactant. You can determine which reactant is limiting by identifying which has the smaller mole-to-mole ratio with the product. This ratio can be found via the coefficients of the balanced reaction.

4 A₂ + 3 B₂ ---> 6 AB

4 moles A₂
------------------ = mole-to-mole ratio A₂/AB
6 moles AB

3 moles B₂
------------------ = mole-to-mole ratio B₂/AB
6 moles AB

Since the mole-to-mole ratio between B₂ and AB is smaller, B₂ must be the limiting reactant.

3 0

2 years ago