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3 years ago

Hydrogen Cyanide, HCN, is a poisonous gas. The

Chemistry

1 answer:

Nikolay [14]3 years ago

5 0

mass of NaCN = 0.539 g

Further explanation

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

2NaCN (s) + H₂SO₄ (aq) ⇒ Na₂S0₄ (aq) + 2HCN (g)

MW HCN = 27,0253 g/mol

MW NaCN = 49,0072 g/mol

Assume there is 1 kg air, and mass of HCN = 300 mg

mol HCN :

mass 300 mg = 0.3 g

$\tt mol=\dfrac{0.3}{27.0253}=0.011$

NaCN : HCN = 2 : 2(mole ratio from equation), so :

mol NaCN = mol HCN = 0.011

mass NaCN :

$\tt mass=mol\times MW\\\\mass=0.011\times 49.0072=0.539~g$

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Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

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The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

3 years ago

Rock samples, taken from the Moon, contain many of the same minerals as those found on Earth. Why might this be the case?

Feliz [49]

The moon used to be connected with Earth. A meteor hit several billion years ago and split the Earth and Moon apart. This explains why they share the Sam properties.

8 0

3 years ago

You’ve just started a new job in the laboratory of the Food and Drug Administration. Your supervisor asks you to prepare a 1.50-

anastassius [24]

Answer: 118.5 grams

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in ml = 500 ml

$1.50=\frac{moles\times 1000}{500ml}$

moles = 0.75

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}$

0.75 = $\frac{\text {given mass}}{158g/mol}$

mass of $KMnO_4$ = 118.5 grams

Thus mass of $KMnO_4$ needed to prepare 500 mL of this solution iis 118.5 grams

6 0

3 years ago

Will mark BRAINLIEST

likoan [24]

Answer:

A and D

Explanation:

4 0

3 years ago

Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima

Rzqust [24]

Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

a) [H⁺] = 1x10⁻¹⁴ / 6x10⁻¹² = 1.67x10⁻³ M. Acidic.

b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

g) [H⁺] = 1x10⁻¹⁴ / 5x10⁻⁵ = 2x10⁻¹⁰ M. Basic.

h) [H⁺] = 1x10⁻¹⁴ / 1x10⁻⁷ = 1x10⁻⁷ M. Neutral.

Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

[OH⁻] = 1x10⁻¹⁴ / 5.2x10⁻⁵

[OH⁻] = 1.92x10⁻¹⁰ M

PArt C:

In this case, we will use expression of part A, and replace the given values:

[H⁺] = 1x10⁻¹⁴ / 2.7x10⁻²

[H⁺] = 3.7x10⁻¹³ M

8 0

3 years ago