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3 years ago

14

In each row, check the box under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g. have t

he larger Ka
a) HCH3CO2 vs HCH3SO2

b) H3PO4 vs H3PO3

c HClO2 vs HClO

1 answer:

Afina-wow [57]3 years ago

3 0

Best guess
Answer: b

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Find the linear function Cequalsf(F) that gives the reading on the Celsius temperature scale corresponding to a reading on the

padilas [110]

Answer:

C = (5/9) F - (160/9)

They both read equal at Z = - 40

Explanation:

We are looking for a linear function so we can write the following condition

Y = aX + b

Applying it to the exercise we got C = a F + b

Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212

0 = 32 a + b (1)

100 = 212 a + b (2)

From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)

and b = - (5/9)32 = - 160/9

Finally the linear function is C = (5/9) F - (160/9)

Both readings are equal at a Z number so

Z = (5/9) Z - 160/9

(4/9) Z = -160/9 and Z = - 40

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3 years ago

What is the most striking part of this simulation?

oee [108]

None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.

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3 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

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3 years ago

At its critical point, ammonia has a density of 0.235 g cm23. You have a special thickwalled glass tube that has a 10.0mm outs

salantis [7]

Answer:

$\large \boxed{\text{69.3 mg}}$

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}$

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3 years ago

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Answer: 2nd option

Explanation: took the quizz

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