- Write equations for (a) the decomposition of carbon monoxide-

When the oxide of generic metal m is heated at 25.0 °c, only a negligible amount of m is produced. o2m(s ----> m(s o2(g delta

Len [333]

(1) MO₂(s) + C(s) → M(s) + CO₂ (g), ΔG₁ = 288.9 kJ/mol
(2) C(s) + O₂(g) → CO₂(g), ΔG₂ = -394.4 kJ/mol
By adding both equations 1 + 2 we get the coupled reaction:
MO₂(s) + 2 C(s) + O₂(g) → M(s) + 2 CO₂(g)
ΔG⁰ = ΔG₁ + ΔG₂
= 288.9 + (-394.4) = -105.5 kJ/mol = -105500 J/mol
Temperature T = 25 + 273.15 = 298.15 K
Molar gas constant R = 8.314 J/mol.K
K = $e^{ \frac{-dG0}{RT} }$
= $e^{ \frac{105500}{8.314 x 298.15} }$
= 3.05 x 10¹⁸

6 0

2 years ago

How many total moles of ions are released when the following sample dissolves completely in water? Enter your answer in scientif

Katena32 [7]

<u>Answer:</u> The number of moles of strontium bicarbonate is $7.5\times 10^{-9}mol$

<u>Explanation:</u>

Formula units are defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is $6.022\times 10^{23}$

We are given:

Number of formula units of $Sr(HCO_3)_2=4.55\times 10^{15}$

As, $6.022\times 10^{23}$ number of formula units are contained in 1 mole of a substance.

So, $4.55\times 10^{15}$ number of formula units will be contained in = $\frac{1}{6.022\times 10^{23}}\times 4.55\times 10^{15}=7.5\times 10^{-9}mol$ of strontium bicarbonate.

Hence, the number of moles of strontium bicarbonate is $7.5\times 10^{-9}mol$

5 0

2 years ago

Which particle has a negative charge?

sweet [91]

Answer: A

Explanation: Beta particles have a charge of -1

6 0

2 years ago

Read 2 more answers

An air mass usually forms in an area of _____. low pressure because it's stable low pressure because it's unstable high pressure

horsena [70]

Answer: high pressure because it's stable

Explanation:

Air mass is volume of air which has stable temperature, humidity and pressure horizontally. Over time, each air mass acquires properties of the region by residing over same part of a surface.

Areas of low pressure and high pressure occur where there is warm air and cold air respectively. An air mass usually forms over an area of high pressure. Warm air rises up and cold air takes its place. Warm air has low density and low pressure where as cold air has high density and pressure and therefore, sinks to the bottom. This is a stable condition. The movement of air mass is responsible for maintenance of temperature conditions on Earth.

6 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.