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3 years ago

What mass of Na2SO4 is needed to make 2.5L of 2.0M solution? (Na=23g; S=32g; O=16g)

2 answers:

8 0

The molar mass of Na₂SO₄ -
2 x Na - 2 x23 = 46
1 x S - 1 x 32 = 32
4 x O - 4 x 16 = 64
total = 46 + 32 + 64 = 142 g/mol
the molarity of solution - 2.0 M
in 1 L of solution , 2.0 moles
Therefore in 2.5 L - 2 mol/L x 2.5 L = 5 mol
then the mass of Na₂SO₄ required = 142 g/mol x 5 mol = 710 g

Mrac [35]3 years ago

3 0

Answer:

710

Explanation:

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For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
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3 years ago

Compare and contrast the concepts of average mass and relative mass. Which one is more accurate, and why? Why is the one you did

Brut [27]

Answer:

Relative and average atomic mass both describe properties of an element related to its different isotopes.

Explanation:However, relative atomic mass is a standardized number that's assumed to be correct under most circumstances, while average atomic mass is only true for a specific sample.

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3 years ago

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In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(a

stiv31 [10]

Answer : The concentration of the NaOH solution is, 0.738 M

Explanation :

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL$

$M_2=0.738M$

Thus, the concentration of the NaOH solution is, 0.738 M

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3 years ago

Jake hangs two balloons from a desk so that they are close together but not touching. Jake rubs one balloon with a wool cloth. H

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The objects that are rubbed together will create charges among themselves due to friction.
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2 years ago

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Which statement describes the Arrhenius interpretation of acids and bases?

Alexeev081 [22]

Answer:

D. It is limited to situations that involve aqueous solutions or specific compounds.

Explanation:

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A is wrong because the Bronsted-Lowry interpretation has a wider range of applications. Bronsted-Lowry acids and bases don't even need to be aqueous, so it is not limited to just aqueous solutions. They include any substance that can donate or accept a H+.

B is wrong because A is wrong. A and B basically say the same thing, that the Arrhenius interpretation has a wider range of applications than the Bronsted-Lowry interpretation.

C is wrong because the definition of an Arrhenius base is any substance that increases the concentration of OH-, or hydroxide ions. C completely counters this statement.

Here's photo for proof incase you're doubtful of my answer & explanation. Please click the heart if it helped.

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2 years ago

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