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scoray [572]
4 years ago
13

1,645.6 expressed in scientific notation equals?

Chemistry
2 answers:
12345 [234]4 years ago
8 0
1.6456 x 10^3 (ten to the third power)
gizmo_the_mogwai [7]4 years ago
6 0

Answer : The correct representation is, 1.6456\times 10^{3}

Explanation :

Scientific notation : It is defined as the method or a way or representation of expressing the given number which are too big or too small that is written in the decimal form. That means, it always written in the power of 10 form.

For example : 8000 is written as, 8\times 10^3

As we are given the digit which is, 1645.6

This number is written in scientific notation as :

1.6456\times 10^{3}

Therefore, the correct representation is, 1.6456\times 10^{3}

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Zn + H₂SO₄ = ZnSO₄ + H₂

Zn⁰ - 2e⁻ = Zn²⁺  zinc is the reducing agent
2H⁺ + 2e⁻ = H₂    hydrogen ions is oxidizer
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Im found a tool in his father’s toolbox and was curious about what it is used for. What does this tool do?
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It measures the length
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In calculating the concentration of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, the stepwise formation constants are as follows: K1=1.90×104
pashok25 [27]

Answer:

Kf = 1.11x10¹³

Explanation:

The value of Kf for a multistep process that involves an equilibrium at each step, is the multiplication of the constant of the equilibrium of each step.

Kf = K1xK2xK3xK4

Kf = 1.90x10⁴ x 3.90x10³ x 1.00x10³ x 1.50x10²

Kf = 1.11x10¹³

4 0
3 years ago
Lead melts at 328 ℃. How much heat is required when 23.0 g of solid lead at 297 K condenses to a liquid at 702 K?
son4ous [18]

Heat = 1.74 kJ

<h3>Further explanation</h3>

Given

melts at 328 ℃ + 273 = 601 K

mass = 23 g = 0.023 kg

initial temperature = 297 K

Final tmperature = 702 K

Required

Heat

Solution

1. raise the temperature(297 to 601 K)

c of lead = 0.130 kJ/kg K

Q = 0.023 x 0.13 x (601-297)

Q = 0.909 kJ

2. phase change(solid to liquid)

Q = m.Lf (melting/freezing)

Q = 0.023 x 23 kj/kg = 0.529 kJ

3. raise the temperature(601 to 702 K)

Q = 0.023 x 0.13 x (702-601)

Q = 0.302 kJ

Total heat = 1.74 kJ

7 0
3 years ago
Determine the ph of a 0. 35 m aqueous solution of CH3NH2 (methylamine). The kb of methylamine is?
guajiro [1.7K]

The pH of 0.35 M of CH₃NH₂ is 12.09 and the k_b for methylamine is 4.4 x 10⁻⁴.

<h3>What is pH?</h3>

pHis the quantitative measure of acidity and basicity of an aqueous or other liquid solution. The scale range goes from 0 to 14. Water has a pH of 7 and is neutral in nature.

<h3>What is dissociation constant?</h3>

The dissociation constant is an equilibrium constant that describes the dissociation or ionization of a base or an acid

k_a describes the dissociation of an acid

k_b describes the dissociation of a base

For methylamine,

CH_3NH_2 + H_2O\Leftrightarrow CH_3NH_3^+ + OH^-

Initial concentration of methylamine = 0.35 M

Initial concentration of products = 0

Let, at equilibrium concentration of CH₃NH₂ = 0.35 - x

Then, concentration of CH₃NH₃⁺ and OH⁻ is x and x respectively

k_b = \frac{[CH_3NH_3^+] [OH^-]}{[CH_3NH_2]}

The dissociation constant for methylamine, k_b = 4.4 x 10⁻⁴

4.4 \times 10^-^4 = \frac{x.x}{0.35} \\x^{2} = 4.4 \times 10^-^4 \times 0.35\\x^{2}  = 1.54  \times 10^-^4 \\

x =0.0124

pOH = -log[OH] = -log(0.0124) = 1.91

pH + pOH = 14

pH = 14 - 1.91 = 12.09

Thus, the pH of methylamine is 12.09 and k_b is 4.4 x 10⁻⁴

Learn more about pH:

brainly.com/question/172153

#SPJ4

7 0
2 years ago
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