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3 years ago

(5.391 × 10−7) + (2.5531 × 10−6)

Chemistry

2 answers:

Alexeev081 [22]3 years ago

6 0

Answer:

66.441

I hope this helps!

Helen [10]3 years ago

6 0

Answer:

66.441

Explanation:

(5.391 × 10-7) + (2.5531 × 10-6)

(46.91) + (19.531)

66.441

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What is a polyprotic buffer?

coldgirl [10]

Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different

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3 years ago

Which formula represents copper(I) oxide?<br> (1) CuO (3) Cu2O<br> (2) CuO2 (4) Cu2O2

faust18 [17]

The answer is (3) Cu2O. Copper (I) has an oxidation state of +1 (that's what the "I" indicates). You can also think of this as copper (I) having a charge of +1. Oxygen has an oxidation state of -2 (that's just a rule you have to know), and you can think of it as oxygen having a charge of -2. You need oxidation numbers in a neutral compound to add up to 0 (or charges in a neutral compond to add up to 0), so you need two Cu to balance the O, which is Cu2O.

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3 years ago

Read 2 more answers

Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

6 0

3 years ago

2C3H7OH + 9O2 --> 6CO2 + 8H2O

Ne4ueva [31]

<h3>Answer:</h3>

733 g CO₂

<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>

[RxN - Balanced] 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given] 5.55 mol C₃H₇OH

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

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})$
Multiply/Divide: $\displaystyle 732.767 \ g \ CO_2$

<u>Step 4: Check</u>

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

732.767 g CO₂ ≈ 733 g CO₂

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3 years ago

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Anna35 [415]

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