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3 years ago

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCI. What volume of 3.00 M HCI

Chemistry

1 answer:

frozen [14]3 years ago

5 0

Answer:

30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.

Explanation:

M1 = 3.00 M

M2 = 1.80 M

V2 = 50 .0 mL = 50 /1000 L = 0.05 L

V1 = unknown

In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.

So we use the equation:

M1 V1 = M2 V2

V1 = M2 V2 / M1

V2 = 1.80 * 0.05 / 3.0

V2 = 0.09 /3.0

V2 = 0.03 L or 30 mL

To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.

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If the reaction is represented by:
PCl₃ + Cl₂ <-> PCl₅ (exothermic)

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Decrease the volume: decrease
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2 years ago

What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?

Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, $T_i=15^{\circ} C$

Final temperature, $T_f=85^{\circ}$

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

$Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ$

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3 years ago

Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2

Serga [27]

Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

(1) H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

-(1) SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

-(1) + (2) + (3): H₂(g) + S(g) → H₂S(g)

ΔH = +519kJ - 242kJ - 297kJ = -20 kJ

I hope it helps!

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2 years ago

Which sentence in a 5-paragraph essay states the essay's main idea?

dimaraw [331]

The main idea of an essay is usually portrayed in the thesis.

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2 years ago

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Oksanka [162]

Answer:

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Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

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You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

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C (−x) (+x) (+x)

E (0.20−x) x x

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