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3 years ago

I don’t understand this one at all

Chemistry

1 answer:

harkovskaia [24]3 years ago

3 0

Answer:

NaF(aq)

Explanation:

When sodium bromide react with fluorine sodium fluoride is formed. Florine is more reactive than bromine that's why it displace the bromine and form compound with sodium.

Chemical reaction:

NaBr(aq) + F₂(aq) → NaF(aq) + Br₂(aq)

Balanced chemical equation:

2NaBr(aq) + F₂(aq) → 2NaF(aq) + Br₂(aq)

Sodium is alkali metal and have one valance electron. It loses one electron and show +1 charge. While all halogen elements need one electron to complete the octet and show -1 charge.

That's why one halogen atom is combine with one sodium atom to make overall compound neutral.

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A sample of limestone (calcium carbonate, CaCO3) is heated at 950 K until it is completely converted to calcium oxide (CaO) and

pav-90 [236]

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

3 0

2 years ago

Consider the reactionI2(g) + Cl2(g)2ICl(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surrou

Georgia [21]

We know,

$\Delta H_{I_2(g)}=62.438\ KJ/mol\\\\\Delta H_{Cl_2(g)}= 0.0\ KJ/mol\\\\\Delta H_{ICl(g)}=17.78\ KJ/mol$

For given reaction, $I_2(g)+Cl_2(g)\ -->\ 2ICl(g)$

$\Delta H_{rxn}=2\Delta H_{ICl(g)}-\Delta H_{I_2(g)}-\Delta H_{Cl_2(g)}\\\\\Delta H_{rxn}=2(17.78)-0-62.438\ KJ/mol\\\\\Delta H_{rxn}=-26.878\ KJ/mol$

For , 2.41 moles of $I_2$ :

$\Delta H_{rxn}=2.41\times (-26.878)\ KJ\\\\\Delta H_{rxn}=-64.78\ KJ$

We know :

$\Delta S = -\dfrac{\Delta H_{rxn}}{T}\\\\\Delta S = -\dfrac{-64.78}{298}\ KJ/K\\\\\Delta S =-0.21738 \ KJ/K\\\\\Delta S=-217.38\ J/K$

Hence, this is the required solution.

7 0

3 years ago

Stoichionetry Quiz

Maru [420]

Answer:

Explanation:

first form your equation and it is C3H8+5O2-4H2O+3CO2 44g of C3H8 produce 3moles of CO2/400g of C3H8 produce x the answer is 27.27moles when you cross multiply

4 0

3 years ago

One application of Hess's Law (which works for ΔH, ΔS, and ΔG) is calculating the overall energy of a reaction using standard en

VikaD [51]

Answer:

The standard change in free energy for the reaction = - 437.5 kj/mole

Explanation:

The standard change in free energy for the reaction:

4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)

Given that ΔGf(KClO3(s)) = -290.9 kJ/mol;

ΔGf(KClO4(s)) = -300.4 kJ/mol;

ΔGf(KCl(s)) = -409 kJ/mol

According to Hess's law

ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)

⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}

= - 901.2 - 409 + 872.7

= - 437.5 kj/mole

3 0

3 years ago

The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12

Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

$120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu$

7 0

3 years ago