CHEMISTRY COLLEGE

Calculate the number of moles in 2.60 grams of SnO2 ? Please show your work to receive credit.

Answers

Answer 1

Answer:

Answer:

There are 0.017252 moles in SnO2

Explanation:

n= m/M

n=2.6/(118.710 + 2(16))

n= 2.6/150.71

n= 0.017252

Remeber to round based on the significant digits in the question.

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3. Theoretically how many grams of magnesium is required to produce to 5.0 g ofMagnesium oxide?
What is the molecular formula of each of the following compounds? Part A Empirical formula CHO2, molar mass = 180.0 g/mol
Topic 1 Homework Homework – Due in 13 hours T2HW Question 16 - Challenge Homework – Unanswered The distance from San Francisco to Los Angeles is approximately 385 miles. You and your friends decide to cycle from San Francisco to Los Angeles. If the distance between the cities is about 385 miles and your doctor tells you that you need to drink 1 L of water for every 1 km that you cycle, how many Lof water will each cyclist need to drink on the journey? Enter your answer as a number using 3 significant figures without units. Do not enter the word "liters" as part of your answer. 1609 m = 1.0 mi Numeric Answer Unanswered
1. Which statement describes the particles of an ideal gas, based on thekinetic molecular theory?* O There are attractive forces between the particles. O The particles move in circular paths. O The collisions between the particles reduce the total energy of the gas. О The volume of the gas particles is negligible compared with the total volume of the gas.

How many electrons does a Bromine (Br) atom have?(Given Information)
Atomic Number : 35
Neutrons: 45
Charge; -1

Answers

35 yooooooooo...........

In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2 g of water are obtained. Determine the percent yield of the reaction.

Answers

The percent yield of the reaction : 89.14%

Further explanation

Reaction of Ammonia and Oxygen in a lab :

4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$(80)/(17)=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt (120)/(32)=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt (4.706)/(4)/ (3.75)/(5)=1.1765/ 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt (6)/(5)* 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=(actual)/(theoretical)* 100\%\n\n\%yield=(72.2)/(81)* 100\%=89.14\%$

Suppose you have been given the task of distilling a mixture of hexane + toluene. Pure hexane has a refractive index of 1.375 and pure toluene has a refractive index of 1.497. You collect a distillate sample which has a refractive index of 1.441. Assuming that the refractive index of the hexane + toluene mixture varies linearly with mole fraction, what is the mole fraction of hexane in your sample?

Answers

Answer:

0.4590

Explanation:

How the refractive index of the hexane + toluene mixture varies linearly with mole fraction, it means that the mole fraction is the fraction that each pure index contribute for the mixture index, so, calling xh the mole fraction of hexane and xt the mole fraction of toluene:

1.375xh + 1.497xt = 1.441

And, xh + xt = 1 (because there are only hexane and toluene in the mixture), so xt = 1- xh

1.375xh + 1.497(1-xh) = 1.441

1.375xh + 1.497 - 1.497xh = 1.441

-0.122xh = -0.056

xh = -0.056/(-0.122)

xh = 0.4590

A chemist prepares a solution of magnesium fluoride MgF2 by measuring out 0.00598μmol of magnesium fluoride into a 50.mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in /μmolL of the chemist's magnesium fluoride solution. Round your answer to 2 significant digits.

Answers

Answer:

0,12 μmol/L of MgF₂

Explanation:

Preparation of solutions is a common work in chemist's life.

In this porblem says that you measure 0,00598 μmol of MgF₂ in 50,0 mL of water and you must calculate concentration in μmol/L

You have 0,00598 μmol but not Liters.

To obtain liters you sholud convert mL to L, knowing 1000mL are 1 L, thus:

50,0 mL (1L/1000mL) = 0,05 L of water.

Thus, concentration in μmol/L is:

0,00598 μmol / 0,05 L = 0,12 μmol/L -The problem request answer with two significant digits-

I hope it helps!

Larry was told that a certain muscle cream was the newest best thing on the market and claims to double a person’s muscle power when used as part of a muscle-building workout. Interested in this product, he buys the special muscle cream and recruits Patrick and SpongeBob to help him with an experiment. Larry develops a special marshmallow weight-lifting program for Patrick and SpongeBob. He meets with them once every day for a period of 2 weeks and keeps track of their results. Before each session Patrick’s arms and back are lathered in the muscle cream, while Sponge Bob’s arms and back are lathered with the regular lotion.Which person is the control group?

SpongeBob

SpongeBob

Patrick

Patrick

Larry

Answers

SpongeBobs twin brother and Larry and Patrick’s twin brother

Final answer:

In the experiment, SpongeBob is the control group because regular lotion is used instead of the muscle cream. This allows a comparison with Patrick(rightly known as the experimental group) who uses the special muscle cream.

Explanation:

In the given scenario, SpongeBob represents the control group. In any experiment, the control group is the one that is kept normal or unchanged to be able to compare the effects of the variable being tested. In this case, it's the use of the special muscle cream. Patrick, whose arms and back are lathered with the muscle cream before each session, represents the experimental group because he is exposed to the variable being tested, which is the muscle cream. On the other hand, SpongeBob, who is given regular lotion instead of the special muscle cream, is part of the control group because he helps to provide a baseline for comparison.

Learn more about Control Group here:

brainly.com/question/32257330

#SPJ3

A cubic box with sides of 20.0 cm contains 2.00 × 1023 molecules of helium with a root-mean-square speed (thermal speed) of 200 m/s. The mass of a helium molecule is 3.40 × 10-27 kg. What is the average pressure exerted by the molecules on the walls of the container? (The Boltzmann constant is 1.38 × 10-23 J/K and the ideal gas constant is R = 8.314 J/mol•K .) (12 pts.)

Answers

Answer:

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Explanation:

Side of the cubic box = s = 20.0 cm

Volume of the box ,V= $s^3$

$V=(20.0 cm)^3=8000 cm^3=8* 10^(-3) m^3$

Root mean square speed of the of helium molecule : 200m/s

The formula used for root mean square speed is:

$\mu=\sqrt{(3kN_AT)/(M)}$

where,

= root mean square speed

k = Boltzmann’s constant = $1.38* 10^(-23)J/K$

T = temperature = 370 K

M = mass helium = $3.40* 10^(-27)kg/mole$

$N_A$ = Avogadro’s number = $6.022* 10^(23)mol^(-1)$

$T=(\mu _(rms)^2* M)/(3kN_A)$

Moles of helium gas = n

Number of helium molecules = N = $2.00* 10^(23)$

N = $N_A* n$

Ideal gas equation:

PV = nRT

Substitution of values of T and n from above :

$PV=(N)/(N_A)* R* (\mu _(rms)^2* M)/(3kN_A)$

$PV=(N* R* \mu ^2* M)/(3k* (N_A)^2)$

$R=k* N_A$

$PV=(N* \mu ^2* M)/(3)$

$P=(2.00* 10^(23)* (200 m/s)^2* 3.40* 10^(-27) kg/mol)/(3* 8* 10^(-3) m^3)$

$P=1133.33 Pa =1.133 kPa$

(1 Pa = 0.001 kPa)

1.133 kPa is the average pressure exerted by the molecules on the walls of the container.

Final answer:

The question asks for the average pressure exerted by helium gas molecules on the walls of a cubic container. Using the equation PV = Nmv^2, we can calculate pressure by substituting the given values for volume, number of molecules, mass of one molecule, and root-mean-square speed.

Explanation:

The question is asking to calculate the average pressure exerted by helium gas molecules on the walls of a cubic container. The important formula relating pressure (P), volume (V), number of molecules (N), mass of a molecule (m), and the square of the rms speed (v2) of the molecules in a gas is:

PV = Nmv2,

First, we need to determine the volume of the container, which is the cube of one side, so V = (20 cm)3 = (0.2 m)3. Inserting the given values into the equation and solving for P gives us the desired answer. Recall that the rms speed is given, so no temperature calculations are needed.

Therefore, using all given data points:

Volume (V) = (0.2 m)3

Number of molecules (N) = 2.00 × 1023

Mass of one helium molecule (m) = 3.40 × 10-27 kg

Root-mean-square speed (vrms) = 200 m/s

By substituting these values, we can find the pressure exerted by the gas. This represents an application of kinetic theory of gases which assumes the behavior of an ideal gas.

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