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AnnyKZ [126]
2 years ago
7

What is the chemical name for Se8O7​

Chemistry
1 answer:
irina1246 [14]2 years ago
3 0

Answer:

Selenium oxide

Explanation:

You might be interested in
8.0 mol AgNO3 reacts with 5.0 mol Zn in
Yakvenalex [24]

Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • AgNO₃: 2 moles
  • Zn: 1 mole
  • Ag: 2 moles
  • Zn(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

amount of moles of AgNO_{3}= \frac{5 moles of Znx2 moles of AgNO_{3}}{1 mole of Zn}

<u><em>amount of moles of AgNO₃= 10 moles </em></u>

But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

#SPJ1

4 0
1 year ago
What is the temperature of water at low latitudes please help me
bixtya [17]
The average temperature of the ocean surface waters is about 17 degrees Celsius (62.6 degrees Fahrenheit).

Hoped that helped
6 0
2 years ago
What is meant when one says that acetic acid is a stronger acid than benzoic acid?
kozerog [31]

Its the other way around actually. Benzoic acid is stronger than acetic acid because the electron-donating inductive effect (+I) by the alkyl group on acetic acid destabilise the conjugate base of acetic acid.

hope this helps :)

6 0
3 years ago
Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict wheth
Elena-2011 [213]

Answer:

\Delta S_{surr} = + 0.32113\: kJ/K

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at <em>constant pressure</em> is given by,

\Delta S_{surr} = \frac{-\Delta H}{T}

\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K}  \right ) = + 0.32113\: kJ/K > 0

<u><em>In the given reaction:</em></u>

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

\therefore \Delta S_{system} <  0

As we know, <em>for a spontaneous process, that the total entropy should be positive.</em>

\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0  

<u>Therefore, at the given temperature,</u>

  • if \Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0 then the given reaction is spontaneous
  • if \Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0 then the given reaction is non-spontaneous
6 0
3 years ago
Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the
fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

<u>b) Br2 is the limiting reactant</u>.

c) There will be formed 86.3 grams of BrF3

d) There will remain <u>0.326 moles of F2 = 12.97 grams</u>

<u />

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

<u>Br2 is the limiting reactant</u>. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = <u>0.326 moles of F2 = 12.97 grams</u>

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = <u>86.3 grams</u>

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7 0
2 years ago
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