Determine the quantity of each atom in a compound

1. T or F? An electric field has strength, but no direction.

Anastasy [175]

1:False, it has both

2:Electric potential is the amount of work done to bring the unit positive charge from infinity to that point, while electric potential energy is the energy that is needed to move a charge against the electric field.

3:Towards

4: Answer should be True

6 0

2 years ago

Need help !!!!! ASAP

Oksi-84 [34.3K]

<h2>Hello!</h2>

The answer is:

The approximate volume is 9.84 L.

$V=9.84L$

Since we are given the numer of moles, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant, which is equal to:

$R=0.082\frac{atm.L}{mol.K}$

So, we are given the information:

$n=1.5mol\\Temperature=300K\\Pressure=3.75atm$

Now, substituting the given information and isolating the Volume from The Ideal Gas Law equation, we have:

$PV=nRT$

$V=\frac{nRT}{P}$

$3.75atm*V=1.5mol*0.082*\frac{atm.L}{mol.K}*300K$

$V=\frac{ 1.5mol*0.082*\frac{atm.L}{mol.K}*300K}{3.75atm}\\\\V=\frac{36.9atm.L}{3.75atm}=9.84L$

So, the approximate volume is 9.84 L.

$V=9.84L$

Have a nice day!

7 0

4 years ago

Use the concept of natural selection to explain why organisms typically adapt over time.

CaHeK987 [17]

Organisms is an environment or habitat for species

6 0

3 years ago

Read 2 more answers

48.0 mL of 1.70 M CuCl2(aq) and 57.0 mL of 0.800 M (NH4)2S(aq) are mixed together to give CuS(s) as a precipitate. The other pro

artcher [175]

Answer:

The concentration of the Cu(II) ion is 0.777M

Explanation:

Step 1: Data given

Volume of 1.70 M CuCl2 = 48.0 mL = 0.0480 L

Volume of 0.800 M (NH4)2S = 57.0 mL = 0.0570 L

Step 2: The balanced equation

CuCl2 (aq) + (NH4)2S (aq) → 2 NH4Cl (aq) + CuS (s)

Step 3: Calculate moles CuCl2

moles CuCl2 = 0.0480 L * 1.70 M=0.0816 moles

Step 4: Calculate moles (NH4)2S

moles (NH4)2S = 0.0570 L * 0.800 M = 0.0456 moles

Step 5: Calculate the limiting reactant

The ratio between CuCl2 and (NH4)2S is 1 : 1 so (NH4)2S is the limiting reactant . IT will completely be consumed (0.0456 moles).

CuCl2 is in excess. There will remain 0.0816 - 0.0456 = 0.0360 moles

Step 6: Calculate moles of CuS

For 1 mol CuCl2 we need 1 mol (NH4)2S to produce 2 moles of NH4Cl and 1 mol CuS

For 0.0456 moles we'll produce 0.0456 moles CuS

Step 7: Calculate moles of Cu(II)ion

There remains 0.0360 moles CuCl2.

There will be 0.0456 moles CuS produced

Total moles Cu^2+ = 0.0816 moles

Step 8: Calculate concentration of Cu(II) ion

Concentration = moles / volume

Concentration = 0.0816 moles / (0.048+0.057)

Concentration = 0.777 M

The concentration of the Cu(II) ion is 0.777M

7 0

4 years ago

Using the periodic table determine the atomic mass of Na2C2O4

Mkey [24]

Masses of atoms are the sum of neutrons and protons. Atomic mass given for the element is the weighted atomic masses of the isotopes depending on the abundance of the isotopes.
Atomic masses of the elements making up Na₂C₂O₄ are as follows;
Na - 22.98 a.m.u
2 atoms of Na - 22.98 x 2 = 45.96
C - 12.01 a.m.u
2 atoms of C - 12.01 x 2 = 24.02
O - 15.99 a.m.u
4 atoms of O - 15.99 x 4 = 63.96
Sum of the atomic masses = 45.96 + 24.02 + 63.96 = 133.94
Mass is 133.94 g/mol

4 0

4 years ago