What is pulled into long wires

Suppose 13.6 g of barium nitrate is dissolved in 300. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final mol

Elis [28]

Answer:

The molarity of barium cation in the solution is 0.173 M

Explanation:

Step 1: The balanced equation

Ba(NO3)2(aq) + Na2CrO4 (aq) → BaCrO4(s) + 2NaNO3(aq)

Step 2: Data given

Mass of Barium nitrate = 13.6 grams

Volume of 0.40M sodium chromate = 300 mL

Step 3: Calculate moles of Ba(NO3)2

Moles = mass / molar mass

Moles = 13.6 grams / 261.34 g/mol

Moles = 0.052 moles

Step 4: Calculate moles of Na2CrO4

Moles = Molarity * Volume

Moles Na2CrO4 = 0.40 * 0.3L

Moles Na2CrO4 = 0.12 moles

Step 5: Calculate limiting reactant

Na2CrO4 is in excess so all of Ba(NO3)2 will be consumed and reacts to form BaCrO4(s) in the form Ba2+

Step 6: Calculate moles of Ba2+

n(Ba2+)=n(BaCrO4) =n(Ba(NO3)2 = 0.0520 moles

Step 7: Calculate molarity of Ba2+

C=n/v so C(Ba2+)=0.0520/0.300 = 0.173 M

The molarity of barium cation in the solution is 0.173 M

4 0

3 years ago

Consider the interaction of two hydrogen ls atomic orbitals of the same phase. Which statements below is an incorrest descriptio

Andreyy89

Answer:

B) The molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital.

Explanation:

When two atoms of hydrogen come close to each other , there is formation of molecular orbital . Due to overlap of 1 s orbital of one and 1 s orbital of another atom , two molecular orbitals are formed . One of these molecular orbital has energy less than 1 s atomic orbital . It is called 1 s sigma bonding molecular orbital . The other molecular orbital has energy more than 1 s atomic orbital . It is called antibonding molecular orbital . Two electrons occupy bonding sigma molecular orbital .

So , the statement that "the molecular orbital formed is lower in energy than a hydrogen 1s atomic orbital " is wrong .

4 0

3 years ago

What is the process that moves sediments called?

Mars2501 [29]

Sediments move one place to another in the process called “ erosion”

6 0

4 years ago

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If 14.5kj of heat were added to 485g of liquid water, how much would its temp increase

OlgaM077 [116]

7.15 degrees C The specific heat capacity of water is 4.1813 J/(g*K). So we have 3 values with the units kJ, g and J/(g*K). We can trivially convert from kJ to J by multiplying by 1000. And we want to get a result with the unit K (degrees Kelvin). So let's do it. First, let's cancel out the g unit by multiplying. 4.1813 J/(g*K) * 485 g = 2027.9305 J/K Now we can cancel out the J unit by dividing. But if we divide by the energy, we'll be left with the reciprocal of K, not K. So instead divide by the J/K unit. So 14500 J / 2027.9305 J/K = 7.150146418 K Rounding to 3 significant figures gives us 7.15 K. And since degrees C and degrees K are the same size, the temperature will increase by 7.15 degrees C

6 0

3 years ago

If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after t

Naya [18.7K]

Answer is: the mass of the excess reactant (ethane) leftover is 90.135 grams.
Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g).
m(C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.

6 0

4 years ago

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What is pulled into long wires​

What is pulled into long wires