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1 month ago

4.1 kg of a plastic, used to make plastic bottles, has a carbon footprint of 6.0 kg of carbon dioxide.

Chemistry

1 answer:

lorasvet [2.7K]1 month ago

4 0

Response:

The carbon footprint associated with a plastic bottle weighing 23.5 g is 34.390 g.

Reasoning:

This can be calculated using a straightforward rule of three. Specifically:

$x = \frac{23.5\,g}{4100\,g}\times 6000\,g$

$x = 34.390\,g$

Hence, for a bottle of mass 23.5 g, the carbon footprint totals 34.390 g.

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n the table below, write the density of each object. Then predict whether the object will float or sink in each of the fluids. W

Anarel [2989]

Answer:

0.5 g/mL----- will float

1.0 g/mL---- will float

2.0 g/mL----- will sink

Explanation:

Objects with a density less than or equal to that of water will float due to having a lower mass, while objects with a density exceeding that of water will sink because their mass is greater than that of water. Thus, objects with a density of 0.5 g/mL and 1.0 g/mL will float since they are less dense than water (1 g/mL), whereas an object with a density of 2.0 g/mL will sink.

3 0

1 month ago

How many 2º alkyl bromides, neglecting stereoisomers, exist with the formula c6h13br?

Anarel [2989]

A secondary alkyl halide would be characterized by having a carbon atom connected to two other carbon atoms, with bromine attached to that carbon.

Therefore, bromo-hexane qualifies as a 2-degree or secondary alkyl halide

5 0

9 days ago

The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constan

eduard [2782]

Response: The rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

Rationale:

Based on the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant when $701K$ = $2.57M^{-1}s^{-1}$

$K_2$ = rate constant when $525K$ =?

$Ea$ = activation energy for the process = $1.5\times 10^2kJ/mol=1.5\times 10^5J/mol$

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 701 K

$T_2$ = final temperature = 525 K

Substituting the provided values into this formula yields:

$\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]$

$K_2=0.0606M^{-1}s^{-1}$

Thus, the rate constant at 525 K is, $0.0606M^{-1}s^{-1}$

8 0

1 month ago

A mass of 5.4 grams of aluminum (al) reacts with an excess of copper (ii) chloride (cucl2) in solution, as shown below. 3cucl2 2

eduard [2782]

Answer:

The mass of copper generated is 19.07g

Explanation:

Let's write out the chemical equation;

3CuCl2 + 2Al → 2AlCl3 + 3Cu

3: 2: 2: 3

After confirming that the reaction is balanced, we can continue.

The problem asks for the mass of Cu produced from 5.4g of Al reacting.

Based on the equation, what is the relation between Al and Cu?

2 moles of Al would yield 3 moles of Cu

Expressing this in mass terms, we find;

mass = number of moles * molar mass

Mass of Aluminium = 2 * 26.98 = 53.98 g

Mass of Copper = 3 * 63.546 = 190.638g

This indicates that;

53.98 g of Al would react to create 190.638g of Cu

So, how much Cu would result from 5.4 g of Al?

This leads us to;

53.98 = 190.638

5.4 = x

By cross multiplication, we determine;

x = (190.638 * 5.4) / 53.98

x = 19.07 g

The mass of copper created is 19.07g

8 0

14 days ago

A solution is prepared by adding 1.43 mol of kcl to 889 g of water. the concentration of kcl is ________ molal. a solution is pr

VMariaS [2998]

A mixture is created by dissolving 1.43 mol of potassium chloride (KCl) in 889 g of water. The concentration of KCl works out to be 1.61 molal.
The amount of KCl is 1.43 mol
Water weighs 889 g
The molality can be calculated using the formula:
molality = moles of solute divided by kilograms of solvent
Since 1 kg equals 1000 g, 889 g is 0.889 kg.

Therefore, m = 1.43/0.889 = 1.61 molal.

7 0

1 month ago