Answer:
yes answer os Na because it's electronic configuration is 1s^2,2s^2,2p^6,3s^1
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.
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
Help help help help help
Convert 3.01 x 10^24 molecules of ammonium sulfate to mass
Mass of ammonium sulfate = 660.7 g
Further explanationGiven
3.01 x 10²⁴ molecules of ammonium sulfate
Required
mass
Solution
The mole is the number of particles(molecules, atoms, ions) contained in a substance
1 mol = 6.02.10²³ particles
Can be formulated
N=n x No
N = number of particles
n = mol
No = Avogadro's = 6.02.10²³
mol ammonium sulfate (NH₄)₂SO₄ :
n = N : No
n = 3.01 x 10²⁴ : 6.02 x 10²³
n = 5
mass ammonium sulfate :
= mol x MW
= 5 x 132,14 g/mol
= 660.7 g
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:
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!
A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the same amount of air is placed in a box with a volume of 5.0m3 at 35∘C? Report your answer with two significant figures.
Answer:
Given:
Initial pressure: [tex]3.65\; \rm atm[/tex].Volume was reduced from [tex]45\; \rm m^{3}[/tex] to [tex]5.0\; \rm m^{3}[/tex].Temperature was raised from [tex]25\; ^\circ \rm C[/tex] to [tex]35\; ^\circ \rm C[/tex].New pressure: approximately [tex]3.4\times 10\; \rm atm[/tex] ([tex]34\; \rm atm[/tex].) (Assuming that the gas is an ideal gas.)
Explanation:
Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:
Reduce the volume of the gas from [tex]45\; \rm m^{3}[/tex] to [tex]5.0\; \rm m^{3}[/tex]. Calculate the new pressure, [tex]P_1[/tex].Raise the temperature of the gas from [tex]25\; ^\circ \rm C[/tex] to [tex]35\; ^\circ \rm C[/tex]. Calculate the final pressure, [tex]P_2[/tex].By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)
For this gas, [tex]V_0 = 45\; \rm m^{3}[/tex] while [tex]V_1 = 5.0\; \rm m^{3}[/tex].
Let [tex]P_0[/tex] denote the pressure of this gas before the volume change ([tex]P_0 = 3.65\; \rm atm[/tex].) Let [tex]P_1[/tex] denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between [tex]P_1\![/tex] and [tex]P_0\![/tex]:
[tex]\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0[/tex].
In other words, because the final volume is [tex](1/9)[/tex] of the initial volume, the final pressure is [tex]9[/tex] times the initial pressure. Therefore:
[tex]\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm[/tex].
On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)
Convert the unit of the temperature of this gas to degrees Kelvins:
[tex]T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K[/tex].
[tex]T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K[/tex].
Let [tex]P_1[/tex] denote the pressure of this gas before this temperature change ([tex]P_1 = 32.85\; \rm atm[/tex].) Let [tex]P_2[/tex] denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at [tex]V_2 = V_1 = 5.0\; \rm m^{3}[/tex].
Apply Amonton's Law to find the ratio between [tex]P_2[/tex] and [tex]P_1[/tex]:
[tex]\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}[/tex].
Calculate [tex]P_2[/tex], the final pressure of this gas:
[tex]\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}[/tex].
In other words, the pressure of this gas after the volume and the temperature changes would be approximately [tex]3.4\times 10\; \rm atm[/tex].
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
What is the chemical formula of tin(IV) chloride pentahydrate?
Answer:
SnCl4 * 5H2O
Explanation:
According to the Bohr model of the atom, which particles are allowed to exist in any one of a number of energy levels?
Answer: the line-emission spectrum of an atom is caused by the energies released when electrons. releases energy of only certain values.
General equilibrium problems. ICE type problems.a. Isopropyl alcohol can dissociate into acetone and hydrogen according to the reaction below.At 179 °C, the equilibrium constant for this dehydrogenation reaction is 0.444. i) If 0.166moles of isopropyl alcohol is placed in a 10 L vessel and heated to 179 °C, what is the partialpressure of acetone when equilibrium is attained
Answer:
Explanation:
In a gaseous reaction mixture partial pressure is proportion to mole of the gas concerned .
Pressure of the reactant gas from gas equation
PV = nRT
P = nRT / V
= .166 x .082 x ( 273+179) / 10
= .615 atm
C₃H₇OH = (CH₃)₂CO + H₂
before reaction moles in terms of pressure
.615 0 0
After reaction
.615 - x x x
.444 = x² / ( .615 - x )
.273 - .444 x = x²
x² + .444 x - .273 = 0
x = .361 atm
So partial pressure of acetone is .361 atm at equilibrium.
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!
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:
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
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.
Learn more about boiling point here:
https://brainly.com/question/23549697
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].
What is the empirical formula for glucose
A)
B)
C)
D)
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
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:
ch4(g) + h2o(g) 3h2(g) + co(g) enthalpy of formation of CH4
Answer:
Kc=[[CO][H2]3[CH4][H2O]
3.90=(0.30)(0.10)3[CH4]×0.02
[CH4]=0.023.90×0.30×(0.10)3=5.85×10−2 M
Thus, the concentration of methane in the mixture is 5.85×10−2 M.
Calculate the [H+]
and pH of a 0.000295 M
butanoic acid solution. Keep in mind that the a
of butanoic acid is 1.52×10−5
[H⁺]=6.696 x 10⁻⁵
pH = 4.174
Further explanationGiven
The concentration of 0.000295 M (2.95 x 10⁻⁴ M) butanoic acid solution
Required
the [H+] and pH
Solution
Butanoic acid is the carboxylic acid group. Carboxylic acids are weak acids
For weak acid :
[tex]\tt [H^+]=\sqrt{Ka.M}[/tex]
Input the value :
[H⁺]=√1.52 x 10⁻⁵ x 2.95 x 10⁻⁴
[H⁺]=6.696 x 10⁻⁵
pH = - log [H⁺]
pH = - log 6.696 x 10⁻⁵
pH = 5 - log 6.696
pH = 4.174
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!
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 ☺️
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)
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 { + }
I need help with this!!!
Answer:
0.73g/cm^3
Explanation:
d=m/v
d=11/15
d=0.73
If energy is conserved will the maximum speed of the pendulum depend on the mass, the length, or both? How?
Answer:
BUDDY YO
Explanation:
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:
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]