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2 years ago

Write balanced chemical equations for:

Chemistry

1 answer:

Wewaii [24]2 years ago

3 0

Answer:

a. C (s)+ 02(g)---->C02(g)

b. CO2(g) +C(s)--->2CO(g)

Explanation:

This reaction is the method of purification of some precious metals such a iron(Fe) in it's ore. a.The carbon(coke) is oxidized to carbon (IV) oxide

b. the carbon dioxide gas is reduced with the same carbon to produce carbon(II) oxide.

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Which is older? Dinosaurs or Grasshoppers?

jok3333 [9.3K]

Answer:

Dinosaur is older.

I hope this helps!

5 0

3 years ago

Read 2 more answers

A certain second-order reaction (B→products) has a rate constant of 1.30×10−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 224 s

soldier1979 [14.2K]

Answer:

$\large\boxed{\large\boxed{0.291M}}$

Explanation:

By definition one <em>half-life</em> is the time to reduce the initial concentration to half.

For a <em>second order reaction </em>the rate law equations are:

$\dfrac{d[B]}{dt}=-k[B]^2$

$\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt$

The <em>half-life</em> equation is:

$t_{1/2}=\dfrac{1}{k[A]_0}$

Thus, substitute the<em> rate constant</em> $1.30\times 10^{-3}M^{-1}\cdot s^{-1}$ and the <em>half-life </em>time <em>224s</em> to find [A]₀:

$224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}$

$[A]_o=0.291M$

7 0

4 years ago

Particles of matter are in motion in

inessss [21]

Answer:

liquid and gases

Explanation:

3 0

3 years ago

Read 2 more answers

Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal

german

Answer:

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

Explanation:

$\Delta T_b=K_b\times m$

$\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}$

where,

$\Delta T_f$ =Elevation in boiling point = $(62.5-61.7)^oC=0.8^oC$

Mass of acenapthalene = 0.515 g

Mass of $CHCl_3$ = 15.0 g = 0.015 kg (1 kg = 1000 g)

$K_b$ = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

$0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}$

$\text{Molar mass of acenapthalene}=155.7875 g/mol$

Let the molecule formula of the Acenapthalene be $C_{6n]H_{5n}$

$6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol$

n = 2.0

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

4 0

4 years ago

The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions a

kolezko [41]

Answer:

Mass PbCl₂ = 50.24g

Mass AgCl = 14.84g

Explanation:

The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:

Ag⁺ + Cl⁻ → AgCl(s)

Pb²⁺ + 2Cl⁻ → PbCl₂(s)

If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:

Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = <em>0.00719X moles of Cl⁻ from PbCl₂</em>

<em />

And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:

(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = <em>0.45409 - 0.00698X moles of Cl⁻ from AgCl</em>

<em />

Moles of Cl⁻ that were added in the KCl solution are:

0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.

<em />

Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)

0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles

0.45409 + 0.00021X = 0.46464

0.00021X = 0.01055

X = 0.01055 / 0.00021

X = 50.24g

As X = Mass PbCl₂

And mass of AgCl = 65.08 - 50.24

6 0

4 years ago