How many molecules are there in 45.0 grams of CH ?
I
How many molecules are there in 150.0 grams of CzHs?
Answer:
There are 1.8021 ⋅ 1024 molecules of CH4 in 48 grams of CH4. To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass. A mole is just a unit of measurement. Avogadro's number is equal to 6.022 ⋅1023 molecules/mole. i think please dont complain to me if its wrong im sorry
Explanation:
Asapp please
The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen,
8.38 % Nitrogen, and 9.57 % Oxygen. If its molar mass is 334.0 g/mol what is its empirical and
molecular formula?
The empirical and molecular formula : C₂₁H₂₂N₂O₂
Further explanationGiven
The percent composition of an unknown substance is 75.42% Carbon, 6.63 % Hydrogen, 8.38 % Nitrogen, and 9.57 % Oxygen
Required
The empirical and molecular formula
Solution
C : 75.42 : 12 = 6.285
H : 6.63 : 1 = 6.63
N : 8.38 : 14 = 0.599
O : 9.57 : 16 = 0.598
Divide by 0.598
C : H : N : O = 10.5 : 11 : 1 : 1 = 21 : 22 : 2 : 2
The empirical formula : C₂₁H₂₂N₂O₂
(C₂₁H₂₂N₂O₂)n = 334
(334)n=334
n = 1
How many grams of water at 0 degree c can be frozen into ice at zero degree c if 55 kj of heat is removed
Answer:
what is that's a question
According to specific heat capacity, 13.09 grams of water at 0 degree Celsius can be frozen into ice at zero degree c if 55 kJ of heat is removed.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT, substitution in formula gives Q/c=m which is m= 55/4.2=13.09 grams.
Thus, 13.09 grams of water at 0 degree Celsius can be frozen into ice at zero degree c if 55 kJ of heat is removed.
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Consider the sodium borohydride reduction of camphor to isoborneol. A reaction was performed in which 1.000 g of camphor was reduced by an excess of sodium borohydride to make 0.661 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.
Answer:
m = 1.0164 g
% = 65.03%
Explanation:
First of all, we need to write the chemical reaction that is taking place here. We have the camphor being reduced to isoborneol:
C₁₀H₁₆O + NaBH₄ -----------> C₁₀H₁₈O + NaBH₂
We have a 1:1 mole ratio between Camphor and isoborneol, so, the moles of the camphor will be the same moles produced of isoborneol.
To get the theorical yield we need to calculate the theorical moles produced of isoborneol, then, the mass and compare it to the given mass. In that way we will get the %yield.
The Molar mass of camphor and isoborneol are:
MM C₁₀H₁₆O = (16*1) + (12*10) + 16 = 152 g/mol
MM C₁₀H₁₈O = (18*1) + (12*10) + 16 = 154 g/mol
The moles of camphor will be:
molesC₁₀H₁₆O = 1 / 152 = 0.0066 moles
The mass produced then of isoborneol should be:
mC₁₀H₁₈O = 0.0066 * 154
mC₁₀H₁₈O = 1.0164 gNow, the %yield would be:
% = (0.661 / 1.0164) * 100
% = 65.03%Hope this helps
Please answer, this is due in 30 minutes
Answer:
0.591 g of magnesium phosphate is the theoretical yield.
Magnesium nitrate is the limiting reactant.
Explanation:
Hello!
In this case, since the balanced reaction turns out:
[tex]3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3[/tex]
Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:
[tex]m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2[/tex]
Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.
However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:
[tex]Y=\frac{actual}{0.591g}*100\%[/tex]
Best regards!
Consider the following unknown solution analysis:
Flame test: red flame Solutions reactions: ammonium carbonate - white precipitate (ppt.) ammonium phosphate - white ppt. Halide test: purple hexane layer The complete chemical name for the unknown compound is______. The correct chemical formula of the compound is_______ .
Answer:
The answer is "Strontium iodide and [tex]SrI_2[/tex]".
Explanation:
In the flames test, two ionic metals give red fire. It was calcium (Ca, which gives a red brick flame) and strontium (Sr, which gives a persistent red flame). Both Ca and Sr respond to insoluble carbonate with ammonium carbonate. Ca and Sr produce insoluble ammonium phosphate phosphates. Even then, the water is reactive with calcium sulfate (CaSO4) whereas the strontium sulfate (SrSO4) is illiquid. Because the metal ion formed 3 precipitates, they can't get Ca, and we have Sr as its metal.
The hexane of its halide gives a violet color which would be typical of iodide (I-).
Strontium iodide is the chemical name of the unknown ionic compound.
Strontium iodide's chemical formula is [tex]SrI_2[/tex].
Both diamond and graphite (i.e. pencil lead) consist of carbon atoms. They are only different in their crystalline structures. One carat or 0.20 g, of a high-quality diamond costs up to $5000, while 25 g of pencil lead may only cost $2. Determine the numbers of C-atom in diamond vs. graphite which can be obtained with $1.
Answer:
Moles of carbon atoms = 3.33 × [tex]10^{-6}[/tex] mol
No. of atoms of C in Diamond = 2.007 × [tex]10^{28}[/tex] atom
Atoms of graphite = 6.27 × [tex]10^{23}[/tex] Atoms
Explanation:
given data
Cost of 0.2g of diamond = $5000
Cost of 25 g of graphite = $ 2
solution
we know cost of 0.2g of diamond is $ 5000 so that for 1$
if buy 1$ = [tex]\frac{0.20}{5000}[/tex]
1$ = 4.0 × [tex]10^{-5}[/tex] g Carbon
and Moles of carbon atoms is express as
Moles of carbon atoms = Given mass of Carbon ÷ atomic mass of C .........1
Moles of carbon atoms = 4.0 × [tex]10^{-5}[/tex] g/ 2.0g
Moles of carbon atoms = 3.33 × [tex]10^{-6}[/tex] mol
and
No. of atoms of C in Diamond = No. of moles × Avogadro NO ..............2
No. of atoms of C in Diamond = 3.33 × [tex]10^{-6}[/tex] mol × 6.022 × [tex]10^{28}[/tex]
No. of atoms of C in Diamond = 2.007 × [tex]10^{28}[/tex] atom
Graphite
and wew have given Cost of 25 g of graphite is $2 so for but 1$ we get
for buy $1 = 25÷2 = 12.5 g Of graphite
Moles of graphite = 12.5÷12 = 1.04 mol
Atoms of graphite = 1.04 × 6.022 × 1023
Atoms of graphite = 6.27 × [tex]10^{23}[/tex] Atoms
7. How much ice is left over if 46 kJ of energy is added to 175 g of ice?
Answer:
[tex]m_{leftover}=37g[/tex]
Explanation:
Hello!
In this case, since the melting process involves the heat added to the system, the heat of fusion and the mass of water:
[tex]Q=m\Delta H_{fus}[/tex]
Thus, as it has been reported that the heat of fusion of ice is 333.6 J/g, we can compute the melted mass of ice as shown below:
[tex]m=\frac{Q}{\Delta H_{fus}} =\frac{46000J}{333.6J/g}\\\\m=138g[/tex]
Thus, the ice leftover results:
[tex]m_{leftover}=175g-138g\\\\m_{leftover}=37g[/tex]
Regards!
What is the chemical formula of tin(IV) chloride pentahydrate?
Answer:
SnCl4 * 5H2O
Explanation:
How many Joules of heat are required to raise the temperature of 20.0 grams of water from 30.0oC to 40.0oC?
Answer:
840 J
Explanation:
c ≈ 4200 J / (kg * °C)
m = 20 g = 0,02 kg
[tex]t_{1}[/tex] = 30 °C
[tex]t_{2}[/tex] = 40 °C
The formula is: Q = c * m * ([tex]t_{2} - t_{1}[/tex])
Calculating:
Q = 4200 * 0,02 * (40 - 30) = 840 (J)
Can somebody PLEASE tell me the empirical formula for br2o6
Answer:
BrO3
Explanation:
The empirical formula is the smallest whole-number ratio, you find the greatest common factor (which is 2, in this case), then divide the subscripts by it.
So:
Br2 / 2 = Br1
O6 / 2 = O3
For each bond, show the direction of polarity by placing a + sign next to the atom expected to have a partial positive charge and a − sign next to the atom expected to have a partial negative charge.
(A) { } Ge-Se { }
(B) { } Ge-Br { }
(C) { } Br-Se { }
Answer:
A) { + } Ge-Se { -}
B) { + } Ge-Br { - }
C) { - } Br-Se { + }
Explanation:
The (-)ive sign shall be placed for the atom with higher electronegativity, while the other atom will be electropositive.
a) Electronegativity of Ge = 2.01
Electronegativity of Se = 2.55
{ + } Ge-Se { -}
b) Electronegativity of Ge = 2.01
Electronegativity of Br = 2.95
{ + } Ge-Br { - }
c) Electronegativity of Br = 2.95
Electronegativity of Se = 2.55
{ - } Br-Se { + }
Is the chemical formula below an Element, Molecule or Compound?
H2SO4
[tex]{ \boxed{ \bold{ \blue{Question}}}}[/tex]
Is H2SO4 an element, compound or molecule?
[tex]{ \boxed{ \bold{ \blue{Answer}}}}[/tex]
H2SO4 (sulphuric acid) is a compound. It is made up of 2 hydrogen atoms, 1 sulphur atom, and 4 oxygen atoms.
_______________________________
[tex] \underline \bold \orange{hope \: it \: helps}[/tex]
A chemical compound is any substance that consists of two or more different elements combined together.
I think that H2SO4 fits this definition of a compound .
HOPE IT HELPS
PLEASE MARK ME BRAINLIEST ☺️
WILL MARK BRANLIEST FOR CORRECT ANSWER! Given the following equation, write the expression for its relative rate.
2N2O(g) — 2N2(g) + O2(9)
[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]
Further explanationReaction
2N2O(g) — 2N2(g) + O2(g)
Required
relative rate
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
so the relative rates for the reaction above are :
[tex]\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}[/tex]
Potassium nitrate (KNO3) is a water-soluble white powder that
is often used as a plant fertilizer. What is the molar
concentration of a solution made up of 505 grams of
potassium nitrate mixed with 250 mL water? The molar mass
of KNO3 is 101.1 g/mol. Round your answer to 3 sig figs.
The molar concentration, often called molarity, describes how much of a substance (a solute) is present per unit of solvent. By definition, the molar concentration (M) is equal to the number of moles (n) of solute divided by the number of liters (the volume, or V) of the solution.
Here, your solute is potassium nitrate, or KNO3. You're given the mass of KNO3 (505 g), but you need to convert this quantity to moles before you can find the molarity. To go from mass to moles, simply divide the mass of the substance by its molar mass (given to you as 101.1 g/mol).
[tex]505 \text{ g KNO}_\text{3} \div 101.1 \text{ g KNO}_3/\text{mol KNO}_{3} = 4.995 \text{ mol KNO}_3[/tex]
Now that we have the moles of solute, we divide by the liters of solution. We're given the volume of solution in milliliters, so to convert to liters, simply divide by 1000 (1 L = 1000 mL, so 1 mL = 1/1000 mL). Our volume of solution is thus 0.250 L.
Finally, we can calculate the molar concentration of the KNO3 solution:
[tex]4.995 \text{ mol KNO}_{3 }\div 0.250 \text{ L} = 19.98 \text{ }\frac{\text{moles}}{\text{liters}}\text{ of KNO}_3[/tex]
But, we're told to round our answer to three sig figs. Thus, our rounded and final answer would be 20.0 moles/liters of KNO3.
Binary compounds are formed by ............... ............... elements.
Answer: i think its A diatomic compound..
Explanation: hope i helped! sorry if im wrong!
Start with 3 breads and 2 cheeses. How many products are made? 1 What are the leftovers?
Answer:
1 product is made. The left overs are 1 piece of bread and 1 slice of cheese
Explanation:
From where do the placenta and umbilical cord develop?
Answer:
it develops from the womb
Answer:
outer cells of the blastocyst
Explanation:
on edg:)
what other traits besides phisical ones could be passed on from parent offspring
Answer:
Love for Music
Explanation:
This is one example of many non-physical traits. In the womb a mother can listen to her favorite music and the growing baby could grow to like it in the womb!
This is just one of the many other traits that could be passed to their offspring.
Hope this Helps!
What is the empirical formula for glucose
A)
B)
C)
D)
Materials
Computer with Internet access 14 Days: A Timeline article
Assignment Instructions
For this project, you are expected to submit two (2) items:
1. Completed Tracking Hurricane Katrina chart
2. Brief essay to summarize your findings about hurricane forecasting and advance warning
systems for hurricanes
Step 1: Prepare for the project.
a) Read through the guide before you begin so you know the expectations for this project.
b) If anything is not clear to you, be sure to ask your teacher.
Step 2: Read the background information about hurricanes.
a) Be sure you can answer these questions:
1.) How do hurricanes form?
2.) How are hurricanes classified?
3.) What kind of damage can hurricanes cause?
Step 3: Read the 14 Days: A Timeline article.
a) As you read, complete the Tracking Hurricane Katrina chart, which can be found at the end of this
document. Not all the dates will have every box filled.
b) Be sure that you note the date from the timeline along with the other information shown in the
chart.
Step 4: Dig deeper thr
Answer:
Here it is (sorry its late)
Explanation:
HELP ME PLS I WILL GIVE BRAINLYEST 1. Is this organism affecting the lives of humans? 2. How is it affecting the lives of humans? 3. What are some ways to prevent the spread of zebra mussels?
Answer:
Yes it is because it effects the food chain in many ways like it takes more time to get rid of them and they eat parasites that other living things need. We can get rid of zebra mussels by removing them little by little or putting animals in the water that eats them.
Explanation:
Answer:
By encouraging boaters to carefully clen, drain and dry their boats before launching them in different bodies of water.
Explanation:
I need help with this!!!
Answer:
0.73g/cm^3
Explanation:
d=m/v
d=11/15
d=0.73
One mole of a metallic oxide reacts with one mole of hydrogen to produce two moles of the pure metal
and one mole of water. 5.00 g of the metallic oxide produces 2.32 g of the metal. What is the metallic
oxide? (Use molar masses)
Answer:
Formulas
3.2 Determining Empirical and Molecular Formulas
Learning Objectives
By the end of this section, you will be able to:
Compute the percent composition of a compound
Determine the empirical formula of a compound
Determine the molecular formula of a compound
The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.
Percent Composition
The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
%H=mass Hmass compound×100%
%C=mass Cmass compound×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
%H=2.5g H10.0g compound×100%=25%
%C=7.5g C10.0g compound×100%=75%
EXAMPLE 3.9
Calculation of Percent Composition
Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?
Solution
To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:
%C=7.34g C12.04g compound×100%=61.0%%H=1.85g H12.04g compound×100%=15.4%%N=2.85g N12.04g compound×100%=23.7%
The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.
Check Your Learning
A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?
ANSWER:
12.1% C, 16.1% O, 71.8% Cl
Determining Percent Composition from Molecular or Empirical Formulas
Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:
%N=14.01amu N17.03amuNH3×100%=82.27%%H=3.024amu H17.03amuNH3×100%=17.76%
This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the
An electric circuit has an expected current of 80 amps An electrician measures the current in the circuit at 120 amps. Which
statement provides a possible explanation for this observation?
There is not enough voltage in the circuit.
The circuit has an extra resistor
A resistor in the circuit is broken
There is not enough electrical energy in the circuit.
What is the boiling point of a solution formed by dissolving 0.75 mol of KCl in 1.00 kg of water?
The boiling point of water generally increases as the amount of impurities (which a solute like KCl technically can be thought of) dissolved increases. This relation can be quantified using the equation,
[tex]\Delta T_b = i \times K_b \times m[/tex]
where [tex]\Delta{T}_{b}[/tex] is the change in the water's boiling point (normally taken to be 100 °C), [tex]i[/tex] is the Van 't Hoff factor (the number of particles a single formula unit of the solute dissociates into in water), [tex]K_b[/tex] is the boiling point elevation constant, and [tex]m[/tex] is the molality (moles of solute/kilogram(s) of solvent) of the solution.
We are forming a solution by dissolving KCl in water. KCl is an electrolyte that, in water, will dissociate into K⁺ and Cl⁻ ions. So, for every formula unit, KCl, we obtain two particles. Thus, the Van 't Hoff factor, or [tex]i[/tex], will be 2.
The molality of the solution can be calculated by dividing the number of moles of KCl by the mass of water in kilograms. Since we have 1.00 kg of water, we would be dividing 0.75 mol KCl by 1, giving us a molality (m) of 0.75 m.
We aren't provided the boiling point elevation constant for water. Several authoritative sources give the value 0.512 °C/m, so we will adopt that as our [tex]K_b[/tex].
Note: m = mol/kg as used in this problem.
Plugging everything in,
[tex]\Delta T_b = i \times K_b \times m \\\Delta T_b = 2 \times 0.512 \text{ } \frac{^oC}{mol/kg} \times 0.75 \text{ } \frac{mol}{kg} \\\Delta T_b = 0.768 \text{ } \mathrm{ ^oC}[/tex]
As you can see, our change in boiling point is positive (the boiling point is elevated), and it is also quite modest. Taking 100 °C to be the boiling point of pure water, the boiling point of our solution would be 100 ⁰C + 0.768 ⁰C, or 100.768 ⁰C.
If we are considering significant figures, then we must give our answer to two significant figures (since 0.75 has two sig figs). We can regard the boiling point of water (100 ⁰C) as a defined value. Since our final answer is a sum, the boiling point of our solution to two significant figures would be 100.77 ⁰C.
Given:
Mol = 0.75Mass = 1.00 kgWe know,
Boiling point constant, Kb = 0.51The molality of the solution will be:
= [tex]\frac{Mole}{Mass}[/tex]
= [tex]\frac{0.75}{1}[/tex]
= [tex]0.75 \ m[/tex]
Now,
→ [tex]T_{solution}-T_{water} = Kb\times m\times i[/tex]
By putting the values, we get
[tex]= 0.51\times 0.75\times 2[/tex]
[tex]= 0.765[/tex]
Boiling point of water = 100°Chence,
Solution's boiling point will be:
→ [tex]T_{solution} = 100+0.765[/tex]
[tex]= 100.765^{\circ} C[/tex]
Thus the above approach is right.
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Convert a speed of 141 mi/h to units of feet per minute show work.
Answer:
12408 feet per minute
Explanation:
Given: Speed is 141 mi/h
To find: speed in units of feet per minute
Solution:
Use the following units to convert the given speed into feet per minute.
1 mile = 5280 foot
1 h = 60 minutes
Therefore,
141 mi/h [tex]=\frac{141(5280)}{60} =12408[/tex] feet per minute.
What is the mass in grams of 1.00 x 10 24 atoms of Mn?
a)91.3 g
b) 123.4 g
c) 1.66 g
d) 166 g
91.2 g Mn
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Reading a Periodic TableAvogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Stoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
[Given] 1.00 × 10²⁴ atoms Mn
Step 2: Identify Conversions
Avogadro's Numer
[PT] Molar Mass of Mn - 54.94 g/mol
Step 3: Convert
[DA] Set up: [tex]\displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})[/tex][DA] Multiply/Divide [Cancel out units]: [tex]\displaystyle 91.2321 \ g \ Mn[/tex]Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
91.2321 g Mn ≈ 91.2 g Mn
What is the density of an object with a mass of 8.7 g and a volume of 8.6 cm??
Answer:
1.01 grams/ cm^3
Explanation:
because that's just how it is
A chemistry student is given 2.00 L of a clear aqueous solution at 43.° C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 25.° C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitate, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.062 kg1) Using only the information above, can you calculate the solubility of X in water at 25 degrees C?2) If yes calculate it. Round answer to 2 significant digits
Answer:
Follows are the solution to the given points:
Explanation:
In part 1:
As described and in the query, they become precipitated whenever the solutions are refrozen to [tex]25^{\circ} \ C[/tex].
Afterward, certain precipitate becomes replaced as well as the remaining water is evaporated, it implies that certain precipitate remained throughout the solution to just the container when the entire balance is evaporated.
The unrecoverable salt precipitates whenever the solvent is cooled at [tex]25^{\circ} \ C[/tex]and the remaining salt dissolves. It dissolved salt remains whenever the water is evaporated because as dissolved salt value is given that results can be achieved.
In part 2:
They have precipitation weight = [tex]0.063\ g[/tex]. They have a [tex]2 \ L[/tex] the solution, they may disregard the volume increases due to its precipitation. The intensity therefore is [tex]\frac{0.063}{2} = 0.0315 \ \frac{g}{L}[/tex]