A rock is found to have a mass of 200 g and a volume of 100 ml. what is the rock's density?

The elaboration likelihood model emphasizes the use of __________ persuasion and __________ persuasion methods. A. direct . . .

Angelina_Jolie [31]

Answer:

It's C, direct and peripheral.

Explanation:

Just took the test

6 0

2 years ago

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In the gold foil experiment, what was observed

natima [27]

Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold Foil Experiment to observe the effect of alpha particles on matter.

8 0

3 years ago

What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7

mariarad [96]

The empirical formula of hydrocarbon is CH2

The molecular formula of the hydrocarbon is C6H12

<u><em>Explanation</em></u>

Hydrocarbon is made up of carbon and hydrogen

<h3><u><em> </em></u>Empirical formula calculation</h3>

Step 1: find the moles CO2 and H2O

moles =mass/molar mass

moles of CO2 = 6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol = 0.15 moles

Step 2: Find the moles ratio of Co2:H2O by diving each mole by smallest mole(0.15)

that is for CO2 = 0.15/0.15 =1

For H2O = 0.15/0.15 =1

therefore the mole ratio of Co2 : H2O = 1:1 which implies that 1 mole of Co2 and 1 mole of H2O is formed during combustion reaction.

From the the law of mass conservation the number of atoms in reactant side must be equal to number of atoms in product side

therefore since there 1 atom of C in product side there must be 1 atom of C in reactant side.

In addition there is 2 H atom in product side which should be the same in reactant side.

From information above the empirical formula is therefore = CH2

Molecular formula calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n = 84 g/mol

n = 6

multiply the each subscript in CH2 by 6

Therefore the molecular formula = C6H12

5 0

3 years ago

How to revise in just 1 whole day for chemistry combined science is it possible to get a good grade

Pavlova-9 [17]

it will be hard, but you can do it. Just study given the materials for the course. Understand enthalpy and entropy, and various types of bonding and you'll be fine.

4 0

3 years ago

What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio

evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

$CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)$

We can write the law of mass action for it:

$Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}$

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change $x$ due to the dissociation extent, we are able to rewrite it as shown below:

$1.8x10^{-5}=\frac{x*x}{0.100-x}$

Thus, via the quadratic equation or solve, we obtain the following solutions:

$x_1=-0.00135M\\x_2=0.00133M$

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0

4 years ago

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