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notsponge [240]

3 years ago

5

Magnesium hydroxide is added to a solution of hydrochloric acid. A reaction occurs and magnesium chloride and water are formed.

Testing the _______ would show that a reaction has occurred.

1 answer:

kodGreya [7K]3 years ago

8 0

Answer:Magnesium (Mg) is a - reactant

Hydrogen (H2) is a - product

magnesium chloride (MgCI2) is a - product

hydrochloric acid (HCI) is a - reactant

Explanation: It’s in my notes

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Name three parts of a plant cell that are not found in an animal cell

lions [1.4K]

The three parts are vacuole , cell wall , chloroplast

4 0

3 years ago

Calculate delta Gº for the following system:

Llana [10]

Answer:

ΔG = - 442.5 KJ/mol

Explanation:

Data Given

delta H = -472 kJ/mol

delta S = -108 J/mol K

So,

delta S = -0.108 J/mol K

delta Gº = ?

Solution:

The answer will be calculated by the following equation for the Gibbs free energy

G = H - TS

Where

G = Gibbs free energy

H = enthalpy of a system (heat

T = temperature

S = entropy

So the change in the Gibbs free energy at constant temperature can be written as

ΔG = ΔH - TΔS . . . . . . (1)

Where

ΔG = Change in Gibb’s free energy

ΔH = Change in enthalpy of a system

ΔS = Change in entropy

if system have standard temperature then

T = 273.15 K

Now,

put values in equation 1

ΔG = (-472 kJ/mol) - 273.15 K (-0.108 KJ/mol K)

ΔG = (-472 kJ/mol) - (-29.5 KJ/mol)

ΔG = -472 kJ/mol + 29.5 KJ/mol

ΔG = - 442.5 KJ/mol

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How many types of tide are there?

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Read 2 more answers

A student titrated 20 ml of 0.410 m hcl with 0.320 m naoh. determine the volume of naoh needed at equivalence point

Aleksandr-060686 [28]

Answer:

25.6mL NaOH

Explanation:

We are given the Molarity of the solution ( $\frac{moles}{liters}$ ) and the volume of the solution (.02L).

By multiplying the two together, we can find the moles of solution that are reacted with HCl.

$moles = \frac{.410 moles}{L} *.02L$

This gives us .0082 moles of HCl.

We then find the moles of NaOH that are needed to react with the HCl using the equation.

$HCl + NaOH = NaCl + H_{2} O$

As HCl and NaCl have a 1:1 ratio, we need .0082 mol of NaOH.

Dividing this value by the Molarity of the solution

$\frac{.0082mol}{.320mol/L}$

Gives us the answer, in Liters (.0256), which we can then divide by 100 convert to mL.

3 0

3 years ago

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g

Tanya [424]

Answer:

(A) $\Delta H^{\circ }_{r}= -144 kJ$

(B) $\Delta H^{\circ }_{r}= - 2552kJ$

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

$1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)$

$1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)$

Now, multiplying equation (a) with 2:

⇒ $N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)$

Then equation b is reversed and multiplied with 2:

$2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)$

Now by adding the equation (a) and equation (b), we get:

⇒ $2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)$

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

$\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}$

$= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ$

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

$B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)$

$2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)$

$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)$

$H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)$

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

$6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)$

$4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)$

$6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)$

$6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)$

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

$\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }$

$= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ$

4 0

3 years ago