The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Charles' law gives the relationship between the volume and the temperature of the gas. Mathematically,
Volume ∝ Temperature
i.e. 
We have, V₁ = 1.6 L, T₁ = 278 K, T₂ = 253, V₂=?
So, the new volume is 1.45 L.
Answer :
The concentration of 
before any titrant added to our starting material is 0.200 M.
The pH based on this ion concentration is 0.698
Explanation :
First we have to calculate the concentration of before any titrant is added to our starting material.
As we are given:
Concentration of HBr = 0.200 M
As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion and bromide ion 
.
As, 1 M of HBr dissociates to give 1 M of
So, 0.200 M of HBr dissociates to give 0.200 M of
Thus, the concentration of before any titrant added to our starting material is 0.200 M.
Now we have to calculate the pH based on this ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Thus, the pH based on this ion concentration is 0.698
Answer:
<h2>The species that furnishes the electrons is called the reducing agent. In this case, the reducing agent is zinc metal.</h2>
Explanation:
<h2>Hopes this helps. Mark as brainlest plz!</h2>
US uses the <span>Fahrenheit temperature scale.</span>
