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2 years ago

How can indicator be used to determine if a solution is an acid or base

Chemistry

1 answer:

dalvyx [7]2 years ago

8 0

Answer:

To tell if something is an acid or a base, you can use a chemical called an indicator. An indicator changes color when it encounters an acid or base. There are many different types of indicators, some that are liquids and others that are concentrated on little strips of "litmus" paper.

Explanation:

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1. Take a look at all the elements in period 3. What do you notice? *

Vedmedyk [2.9K]

Answer:

Explanation:

beacause all the elements in period 3 has different valence electrons hence you look at the first element in period 3 is sodium which has a valence electron of 1 when you go across the period to the right their valence electron decreases to be negative

5 0

2 years ago

Read 2 more answers

The airbags that protect people in car crashes are inflated by the extremely rapid decomposition of sodium azide, which produces

Oxana [17]

Answer:

1. NaN₃(s) → Na(s) + 1.5 N₂(g)

2. 79.3g

Explanation:

1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.

NaN₃(s) → Na(s) + 1.5 N₂(g)

2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.

First, we have to calculate the moles of N₂ from the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1atm.(43.0L)}{(0.08206atm.L/mol.K).286.2K} =1.83mol$

The moles of NaN₃ are:

$1.83molN_{2}.\frac{1molNaN_{3}}{1.5molN_{2}} =1.22molNaN_{3}$

The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:

$1.22mol.\frac{65.01g}{mol} =79.3g$

5 0

2 years ago

The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.

Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

$\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]$

Where,

$\Delta H_v$ is the Heat of vaoprization (J/mol)

$T_b$ is the normal boiling point of the gas (K)

$T_c$ is the Critical temperature of the gas (K)

$P_c$ is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

$T_b=307.4\ K$

$T_c=466.7\ K$

$P_c=36.4\ bar$

Applying the above equation to find heat of vaporization as:

$\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]$