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3 years ago

Formula for the compound Tetraoxygen nonabromide?

Chemistry

1 answer:

Vikentia [17]3 years ago

3 0

Answer:

O4Br9

Explanation:

Tetra means 4 and nona means 9. Just use these prefixes to decide how many of each element you have.

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Name a metal and non metal which are liquid at normal pressure and temperature<br>

liraira [26]

Answer:

metal : Mercury(Hg)

non metal : bromine (Br)

Explanation:

mercury is liquid at room temperature and pressure and the same as bromine

6 0

2 years ago

Question 8 of 8> 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a s

Nesterboy [21]

Answer:

12.29 M

Explanation:

The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

Now let's calculate the <u>moles of H₂SO₄ that were titrated</u>:

$0.1284 M * 0.1915L*\frac{1molAcid}{2molNaOH}$ = 0.01229 mol H₂SO₄.

Thus, the <u>concentration of the diluted solution is</u>:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

Finally, the <u>concentration of the original acid solution is:</u>

$1.229 M*\frac{250mL}{25mL}$ = 12.29 M

5 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Describe the day of June 21 at the North Pole in terms of daylight and darkness <br> ( science )

frosja888 [35]

On June 21, as seen from the North pole ...

-- the sun has been up, and it's been light outside,
for the past three months ... ever since March 21 .

-- The sun won't set, and it won't be dark outside,
for another three months ... until September 21.

-- Here at the North pole, it stays daylight for six months straight.
Today, on June 21, we're exactly halfway through the period of
continuous daylight.

7 0

3 years ago

Use the reaction 1₂(s) = 12(g), AH = 62.4 kJ/mol, AS = 0.145 kJ/(mol-K), for

Sati [7]

The Reaction is spontaneous when temperature is 430 K. Hence, Option (C) is correct.

<h3></h3><h3>What is Spontaneous reaction ?</h3>

Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system.

When both of these conditions are met, the reaction occurs naturally.

Spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring.

According to Gibb's equation:

ΔG = ΔH - TΔS

ΔG = Gibbs free energy

ΔH = enthalpy change = +62.4 kJ/mol

ΔS = entropy change = +0.145 kJ/molK

T = temperature in Kelvin

ΔG = +ve, reaction is non spontaneous
ΔG = -ve, reaction is spontaneous
ΔG = 0, reaction is in equilibrium

ΔH - TΔS = 0 for reaction to be spontaneous

T = ΔH / ΔS

Here,

T = 500K

Thus the Reaction is spontaneous when temperature is 500 K.

Learn more about Gibbs free energy here ;

https://brainly.in/question/13372282

#SPJ1

3 0

2 years ago