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nalin [4]
3 years ago
9

____B₂Br₆ + ___HNO₃ ---->___B(NO₃)₃ + ____HBr balance them thanks!!!

Chemistry
1 answer:
natita [175]3 years ago
6 0
1, 6, 2, 6
In the order you wrote them in
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Write a balanced equation for the complete oxidation reaction that occurs when glucose (C6H12O6) reacts with oxygen. Use the sma
Stels [109]

Answer:

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

Explanation:

Glucose (C₆H₁₂O₆) react with oxygen (O₂) to produce carbon dioxide (CO₂) and water (H₂O).

The equation can be written as follow:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

The above equation can be balance as illustrated below:

C₆H₁₂O₆ + O₂ —> CO₂ + H₂O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 6 in front of CO₂ as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + H₂O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H₂O as shown below:

C₆H₁₂O₆ + O₂ —> 6CO₂ + 6H₂O

There are a total of 8 atoms of O on the left side and a total of 18 atoms on the right side. It can be balance by 6 in front of O₂ as shown below:

C₆H₁₂O₆ + 6O₂ —> 6CO₂ + 6H₂O

Now, the equation is balanced.

7 0
3 years ago
Pls help asap pls !!
klio [65]

Answer:

basic solution.

Explanation:

has a pH of 12.

it turns clear indicator pink.

it contains hydroxide ions which conduct electricity

3 0
3 years ago
I need help with this question please someone tell me the answer for this
Andrej [43]
Aueyeowosy yo this is my answer

5 0
2 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
A substance has a boiling point of 78 °C. Which of the following is true about the substance? (5 points) a It will also change f
vovangra [49]

Answer: If a substance has a boiling point of 78^{o}C then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.

Explanation:

The temperature at which vapor pressure of a liquid substance becomes equal to the atmospheric pressure is called boiling point of substance.

At the boiling point, liquid phase and vapor phase remains in equilibrium.

This means that as liquid phase changes into vapor phase and also vapor phase changes into liquid phase at the boiling point.

Thus, we can conclude that if a substance has a boiling point of 78^{o}C then it is true that it will also change from a gas to a liquid at 78 °C while the gas loses energy.

6 0
2 years ago
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