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NikAS [45]
1 year ago
9

Part E

Chemistry
1 answer:
lbvjy [14]1 year ago
4 0

The production of ammonia will be favored at high pressure and low temperature.

<h3>What is an exothermic reaction?</h3>

The term exothermic reaction has to do with a reaction in which the forward reaction is favored at lower temperatures.

Now looking at the reaction coordinate and the equation of the reaction, we know that the production of ammonia will be favored at high pressure and low temperature.

Learn more about ammonia:brainly.com/question/12276882

#SPJ1

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Given a certain quantity of reactant, you calculate that a particular
zlopas [31]

Answer:

Percent yield =  114.5%

Explanation:

Given data:

Calculated amount of product/ theoretical yield = 55 g

Actual amount of product after performing reaction = 63 g

Percent yield = ?

Solution:

Formula:

Percent yield = ( actual yield / theoretical yield )× 100

by putting values,

Percent yield = (63 g/ 55 g)× 100

Percent yield =  114.5%

4 0
3 years ago
What mass of Na2SO4is needed to make 2.5 L of 2.0 Msolution? (Na = 23 g; S = 32 g; O = 16 g)
ivanzaharov [21]

Answer:

mass (g) needed = 710.2 grams Na₂SO₄(s)

Explanation:

Needed is 2.5 Liters of 2.0M Na₂SO₄; formula wt Na₂SO₄ = 142.04g/mol.

mass (grams) of Na₂SO₄(s) = Molarity needed x Volume needed in Liters x Formula Wt of solute

mass (grams) of Na₂SO₄(s) = (2.5L)(2.0M)(142.04g/mol) = 710.2 grams Na₂SO₄(s)

Mixing: Transfer 710.4 grams Na₂SO₄ into mixing vessel and add water-solvent up to but not to exceed 2.5 Liters total volume. Mix until dissolved.

Gives 2.5 Liters of 2.0M Na₂SO₄(aq) solution.

5 0
2 years ago
A 67.6 g sample of zinc is heated, then placed in a calorimeter containing 65.0 g of water. Temperature of water increases from
alexira [117]

Answer:

Hi Im an online tutor and i can assist you with all your assignments. We have experts in all fields. check out our website https://toplivewriters.com/

Explanation:

7 0
2 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

                        = -1839.6 KJ



5 0
3 years ago
What is food nutrients​
Illusion [34]

Answer:

Nutrients arw compounds in foods essential to life and heath

5 0
3 years ago
Read 2 more answers
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