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3 years ago

What dose iff mean when used in a biconditional statement

Chemistry

1 answer:

katovenus [111]3 years ago

4 0

Hello!

A biconditional statement is when one conditional statement and it's converse is true.

A conditional statement is an "If hypothesis, then conclusion" statement.

A converse statement is an "If conclusion, then hypothesis" statement.

Looking at this, both of these statements have their hypothesis and conclusion switched.

Thus, a biconditional statement is a "Hypothesis IFF conclusion." In this statement, IFF means if and only if.

Therefore, when IFF is used in a biconditional statement, it means if and only if.

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And

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3 years ago

Read 2 more answers

Determine the pH of a 5x10^-4 M solution of Ca(OH)2

miss Akunina [59]

Ca(OH)₂ ==> Ca²⁺ + 2 OH-

Ca(OH)₂ is strong Bases

Therefore, the [OH-] equals 5 x 10⁻⁴ M. For every Ca(OH)₂ you produce 2 OH⁻.

pOH = - log[ OH⁻]

pOH = - log [ 5 x 10⁻⁴ ]

pOH = 3.30

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pH + 3.30 = 14

pH = 14 - 3.30

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3 years ago

Ammonia gas decomposes to form nitrogen and hydrogen gases. NH3(g) → N2(g) 3H2(g) If the nitrogen gas is collected in a rigid 2

sveticcg [70]

Decomposition is a chemical reaction that breaks the reactant into two or more products. Moles of nitrogen gas $(\rm N_{2})$ in the cylinder is 1.63 moles.

<h3>What is the ideal gas equation?</h3>

The ideal gas equation states the relation of the hypothetical ideal gas according to the pressure, volume, temperature and moles of the gas. It is given by,

$\rm PV = nRT$

Where,

Pressure (P) = 2000 kPa

Volume (V) = 2L

Temperature (T) = 295 K

Gas constant (R)= 0.08206

Substituting values in the equation:

$\begin{aligned} \rm n &= \rm \dfrac{PV}{RT}\\\\&= \dfrac{2000 \times (\dfrac{1}{101.325}) \times 2}{0.08206 \times 295}\\\\&= 1.63\;\rm mol\end{aligned}$

Therefore, 1.63 moles are produced.

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2 years ago

The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur

anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × $10^{7}$

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × $10^{7}$

ΔG° = -2494.2㏑ 1.7 × $10^{7}$ = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

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1 year ago

What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

Morgarella [4.7K]

This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5% as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

$\% H=\frac{m_H*1}{M_{HF}}*100\%$

Where $m_H$ stands for the atomic mass of hydrogen and $M_{HF}$ for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

$\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%$

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2 years ago