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denis-greek [22]

3 years ago

11

A _______ reaction is when a compound containing carbon and hydrogen (and sometimes oxygen) combines with oxygen gas to produce

carbon dioxide and water.

1 answer:

ch4aika [34]3 years ago

6 0

This is categorized as a combustion reaction.

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Which substance have the same number of carbon (C) atoms?

Vladimir79 [104]

Answer:

Carbon monoxide

Explanation:

Carbon dioxide and carbon monoxide have the same number of carbon atoms.

Mark Brainliest

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3 years ago

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Why are alloys preferable at times?

andriy [413]

Alloys are preferable at times because they offer interesting properties than pure metal. Alloys are made in order to modify or enhance properties, especially mechanical properties. Examples are stainless steel, brass and wrought iron.

3 0

3 years ago

SOMEONE PLEASE HELP ME!!!

pantera1 [17]

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4 years ago

In a similar experiment, a current of 2.15 amps ran for 8 minutes and 24 seconds. The temperature of the water was 26.0°C. The v

lord [1]

Answer:

$\boxed{6.08 \times 10^{23}}$

Explanation:

Data:

I = 2.15 A

t = 8 min 24 s

T = 26.0 °C

V = 65.4 mL

p = 774.2 To

1. Write the equation for the half-reaction

2H₂O ⟶ O₂ + 4H⁺ + 4e⁻

2. Calculate the moles of oxygen

$p = \text{774.2 To} \times \dfrac{\text{1 atm}}{\text{760 To}} = \text{1.0187 atm}$

V = 0.0654 L

T = (26.0 + 273.15) K = 299.15 K

$\begin{array}{rcl}pV& = & nRT\\1.0189 \times 0.0654 & = & n \times 0.08206 \times 299.15\\0.06662 & = & 24.55n\\\\n & = & \dfrac{0.06662}{24.55}\\\\n & = & 2.714 \times 10^{-3}\\\end{array}$

3. Calculate the moles of electrons

$\text{n} = \text{2.714 $\times 10^{-3}$ mol oxygen} \times \dfrac{\text{4 mol electrons}}{\text{ 1 mol ozygen}} = \text{ 0.01086 mol electrons}$

4. Calculate the number of coulombs

t = 8 min 24 s =504 s

Q = It = 504 s × 2.10 C·s⁻¹= 1058 C

5. Calculate the number of electrons

$\text{No. of electrons} = \text{1058 C} \times \dfrac{\text{1 electron}}{1.602 \times 10^{-19}\text{ C}} = 6.607 \times 10^{21}\text{ electrons}$

6. Calculate Avogadro's number

$N_{\text{A}} = \dfrac{6.607 \times 10^{21}}{0.01086} = \mathbf{6.08 \times 10^{23}}\\\\\text{The experimental value of Avogadro's number is } \boxed{\mathbf{6.08 \times 10^{23}}}$

3 0

4 years ago

g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i

kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is $\alpha_{Y^{-4}}$ =0.81

So the Conditional Formation constant= $K_f$ =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL

V e= 25.00 mL

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.

$[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M$

${K^'}_f$

Ca²⁺ + Y⁴ ⇄ CaY²⁻

Initial 0 0 0.001667

change +x +x -x

equilibrium x x 0.001667 - x

${K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\$

$x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14$

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M

pca = —log(2 15x101= 6.7

3 0

4 years ago