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podryga [215]
3 years ago
7

Describe how a mole ratio is correctly expressed when it is used to solve a stoichiometry problem

Chemistry
1 answer:
notka56 [123]3 years ago
3 0
Basically stoichiometry is the measurement of elements, that is, the study of chemical quantities consumed or produced in a chemical reaction. When we are performing it we are using a special chemical counting unit: the mole, a unit of measurement, and one mole of a substance contains 6.022 * 10^23 particles. Now mole ratio is defined as the ratio of moles of one substance to the moles of another substance in a balanced equation. <span>If we are looking for the mole ratio between two substances, we need to look at the balanced equations</span> for the coefficients in front of the substances you are interested in.
This should be your guiding mantra for doing stoichiometry problems!
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prisoha [69]

The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.

<h3>What is equilibrium?</h3>

Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.

A very high value of K indicates that at equilibrium most of the reactants are converted into products.

The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.

When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.

This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.

Hence, option A is correct.

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5 0
3 years ago
What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
3 years ago
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