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4 years ago

A type of men's cologne

Chemistry

1 answer:

Whitepunk [10]4 years ago

8 0

Hey there,

Answer:

<span>Fierce by Abercrombie & Fitch.

Hope this helps :D

<em>~Top♥</em>
</span>

You might be interested in

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 3.30kg of water at 23.8°C. Dur

Brums [2.3K]

Answer:

$T_2= 31.9\°C$

Explanation:

Hello there!

In this case, it is possible to propose an energy balance in order to illustrate how the heat released by the reaction is absorbed by the water:

$-Q_{rxn}=Q_{water}$

Thus, since the heat released by the reaction is -112 kJ (-112000 J), it is possible to define the hear absorbed by the water in terms of mass, specific heat and temperature change:

$-(-112000J)=m_{water}C_{water}(T_2-T_1)$

In such a way, it is possible to define the final temperature as shown below:

$T_2=23.8\°C+\frac{112000J}{3300g*4.18\frac{J}{g\°C} }\\\\T_2= 31.9\°C$

Best regards!

4 0

3 years ago

How many grams of CO2 are used when 7 g of oxygen are produced

lara [203]

308 grams of co2 is used

4 0

3 years ago

Ethylene produced by fermentation has a specific gravity of 0.787 at 25 degree Celsius. What is the volume of 125g of ethanol at

WITCHER [35]

<u>Answer:</u> The volume of given amount of ethanol at this temperature is 159.44 mL

<u>Explanation:</u>

Specific gravity is given by the formula:

$\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}$

We are given:

Density of water = 0.997 g/mL

Specific gravity of ethanol = 0.787

Putting values in above equation, we get:

$0.787=\frac{\text{Density of a substance}}{0.997g/mL}\\\\\text{Density of a substance}=(0.787\times 0.997g/mL)=0.784g/mL$

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$ ......(1)

Given values:

Mass of ethanol = 125 g

Density of ethanol = 0.784 g/mL

Putting values in equation 1, we get:

$\text{Volume of ethanol}=\frac{125g}{0.784g/mL}=159.44mL$

Hence, the volume of given amount of ethanol at this temperature is 159.44 mL

6 0

3 years ago

What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?

AlexFokin [52]

This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

$M=\frac{mol\ solute}{Volume\ solution \ in \ L}$

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

$mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L$

Finally, we plug in these numbers in the equation for the calculation of molarity:

$M=\frac{0.0313mol}{0.100L}\\\\M=0.313M$

Learn more about molarity: brainly.com/question/10053901

8 0

2 years ago

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

coldgirl [10]

Answer:

$\rm_{90}^{231}\text{Th}$

Explanation:

The unbalanced nuclear equation is

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} + X$

Let's write X as a nuclear symbol.

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} + _{Z}^{A}\text{X}$

The main point to remember in balancing nuclear equations is that the sums of the superscripts and of the subscripts must be the same on each side of the reaction arrow.

Then

235 = 4 + A , so A = 235 - 4 = 231, and

92 = 2 + Z , so Z = 92 - 2 = 90

And your nuclear equation becomes

$\rm _{92}^{235}\text{U} \longrightarrow \, _{2}^{4}\text{He} +\, _{90}^{231}\text{X}$

Element 90 is thorium, so

$\rm X = _{90}^{231}\text{Th}$

7 0

4 years ago