All categories

3 years ago

TRUE OR FALSE? An exothermic reaction could be written as: heat + A + B -----> AB

2 answers:

7 0

The correct answer for this statement would be FALSE. It is considered false since exothermic reactions release energy, therefore, this energy can be considered more of a product, instead of the reactant. Hope this is the answer that you are looking for.

IgorC [24]3 years ago

6 0

Answer:

False,the given representation of an exothermic reaction is wrong.

Explanation:

Exothermic reaction are those reaction which releases energy in the form of heat into their surroundings. This release of energy leads to rise in temperature of the of surrounding.

$A+B\rightarrow AB+Heat$

Endothermic reaction are those reaction which energy is absorbed from the surroundings. This absorption of energy leads to lowering in temperature of the of surrounding.

$A+B+Heat\rightarrow AB$

You might be interested in

An ultraviolet ray of light has a wavelength of 1.5x10^-8 m. What is the ray's<br> frequency?

frutty [35]

Answer:

2 E16 Hz or 2 * 10^16 Hz

Explanation:

The formula to determine frequency is f = c / λ.

f = frequency

c = speed of light

λ = wavelength

f = 3E8 / 1.5E-8

f = 2E16

This makes sense because UV light exists roughly

between 8E14 Hz and 3E16 Hz ----- 2E16 Hz falls in that range

8 0

3 years ago

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

Jon is running for city council and wants to campaign on human development goals for his community. Which issue should he use? A

kotegsom [21]

Answer:

E. better schools

Explanation:

This contributes the most to human development in his community because it provides children with more resources and better opportunities than a sub-par education system. If a community is able to better their school system, they will begin to see a lot of positive changes in their area.

6 0

3 years ago

Read 2 more answers

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

3 years ago

The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is ________.

Naddika [18.5K]

Answer:

12.6.

Explanation:

We should calculate the no. of millimoles of KOH and HCl:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

It is clear that the no. of millimoles of KOH is higher than that of HCl:

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

4 0

3 years ago