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3 years ago

Persamaan ion for Mg+HCI

Chemistry

1 answer:

Alenkinab [10]3 years ago

3 0

In the reaction we gives the magnesium chloride and released hydrogen gas.

The reaction is carried out by contaning positive and negative charged on elements as follows

Mg + HCl --->MgCl2 + H2

Net reaction gives

Mg+ + 2Cl- + 2H+---> Mg+² + 2Cl- + 2H

Check it Mg(s) and Mg2+ not equal . The donor is element while the acceptor is an ion. The same process is done for H2 (g) and H+.

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Rain form from clouds, while mist forms at Earth’s surface

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3 years ago

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How to name the hydrocarbons? according to IUPAC i guess?

jeka94

Answer:

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3 years ago

A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C.What

Pachacha [2.7K]

Answer:

0.3832

Explanation:

Givens

m = 437.2 grams

c = ?

delta t = 69.8 - 19.3

H = 8460 J

Formula

H = m*c*Δt

Solution

8460 = 437.2 * c * (69.8 - 19.3)

8460 = 437.2 * c * 50.5

8460 = 22078.6 * c

c = 8460 / 22078,6

c = .3832 J / (oC * gr)

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3 years ago

I need some help with a long chemistry problem. Anything is appreciated!

Mkey [24]

Answer:

The CSI is wrong.

Explanation:

1. Find the volume of the pool

The formula for the volume of a cylinder is V = πr²h .

D = 12 m; h = 10 m

r = D/2 = (12 m/2) = 6.0 m

V = πr²h = π × (6.0 m)² × 10 m = π × 36 m²× 10 m = 360π m³ = 1100 m³

= 1.1. × 10⁶ L

2. Calculate the moles of OH⁻

n = cV = 1.0 × 10⁻² mol·L⁻¹ × 1.3 × 10⁶ L = 11 000 mol of OH⁻

3. Calculate the moles of acetic acid needed for neutralization

HA + OH⁻ ⟶ A⁻ + H₂O

The molar ratio of is 1 mol HA:1 mol OH⁻, so you need 11 000 moL of acetic acid.

4. Calculate the actual moles of acetic acid

You have four 5 L jugs of acetic acid pH 2 .

Volume = 20 L

[H⁺] = 10⁻² mol·L⁻¹ = 0.01 mol·L⁻¹

(a) Set up an ICE table

HA + H₂O ⇌ A⁻ + H₃O⁺

I/mol·L⁻¹: c 0 0

I/mol·L⁻¹: - 0.01 +0.01 +0.01

I/mol·L⁻¹: c - 0.01 0.01 0.01

$K_{\text{a}} = \dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}} = 1.76 \times 10^{-5}$

(b) Calculate the concentration of acetic acid

$\begin{array}{rcl}\dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}}& = & 1.76 \times 10^{-5}\\\\\dfrac{0.01\times 0.01}{c}& = & 1.76 \times 10^{-5}\\\\1 \times 10^{-4} & = & 1.76 \times 10^{-5}c\\c & = & \dfrac{1 \times 10^{-4}}{1.76 \times 10^{-5}}\\\\ & = & \text{6 mol/L}\\\end{array}$