Answer:
Haematology is the specialty important for the diagnosis and management of a wide range of benign and malignant disorders of the red and white blood cells, platelets and the coagulation system in adults and children.
The mass of a crucible and lid is 23.422 g. After adding a sample of hydrate compound the crucible, cover, and contents weigh 24.746 g. After heating with a Bunsen burner to remove the water of hydration, the mass of the crucible, cover, and sample was 24.213 g. How many moles of water did the hydrate compound contain
Answer:
0.030 mole
Explanation:
Mass of crucible + lid = 23.422 g
Mass of crucible + lid + compound = 24.746 g
Mass of crucible + lid + compound - water = 24.213
Mass of water = Mass of crucible + lid + compound + heat
= 24.746 - 24.213
= 0.533 g
Mole of water in the hydrated compound = mass of water in the compound/molar mass of water
= 0.533/18
= 0.0296 mole = 0.030 mole
What is it?? plzzzz help
The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chloride dissolves in water. Be sure to specify states such as (aq) or (s).
Answer:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Explanation:
Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
If you reacted 88.9 g of ammonia with excess oxygen, what mass of water would you expect to make? You will need to balance the equation first.
NH3(g) + O2(g) -> NO(g) + H2O(g)
Explanation:
here's the answer to your question
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
Convert 192 grams of phosphorus pentabromide to molecules.
Convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules.
Answer:
1) 2.69 * 10²³ PBr₅
2) 6.02 * 10²⁴ C₁₂H₂₂O₁₁
Explanation:
Question 1)
We want to convert 192 grams of phosphorus pentabromide to molecules. Note that 192 is three significant figures.
Phosphorus pentabromide is given by PBr₅.
To convert from grams to molecules, we can convert from grams to moles first, and then from moles to molecules.
To convert from grams to moles, we will find the molar mass of PBr₅.
Since the molar mass of P is 30.974 g/mol and the molar mass of Br is 79.904 g/mol, the molar mass of PBr₅ is:
[tex](30.974)+5(79.904) = 430.494\text{ g/mol}[/tex]
And since we want to convert from grams to moles, we can write the following ratio:
[tex]\displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}[/tex]
Where grams is in the denominator, which allows us to cancel them out, leaving us with only moles.
To convert from moles to molecules, we can use the definition of the mole: a mole of one substance has 6.022 * 10²³ amount of that substance.
So, a mole of PBr₅ has 6.022 * 10²³ molecules of PBr₅. Since we want to cancel out the moles, we can write the ratio:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
In combination, starting with 192 grams of PBr₅, we will acquire:
[tex]\displaystyle 192\text{ g PBr$_5$} \cdot \displaystyle \frac{1 \text{ mol PBr$_5$}}{430.494\text{ g PBr$_5$}}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1\text{ mol PBr$_5$}}[/tex]
Cancel like units:
[tex]\displaystyle = 192 \cdot \displaystyle \frac{1 }{430.494}\cdot \displaystyle \frac{6.022 \times 10^{23} \text{ PBr$_5$}}{1}[/tex]
Multiply. Hence:
[tex]=2.6858...\times 10^{23}\text{ PBr$_5$}[/tex]
Since the final answer should have three significant digits, our final answer is:
[tex]= 2.69\times 10^{23} \text{ PBr$_5$}[/tex]
So, there are about 2.69 * 10²³ molecules of PBr₅ in 192 grams of the substance.
Question 2)
We want to convert 3.42 kilograms of table sugar (C₁₂H₂₂O₁₁) to molecules. Note that this is three significant figures.
3.42 kilograms is equivalent to 3420 grams of table sugar.
Again, we can convert from grams to moles, and then from moles to molecules.
First, we will find the molar mass of table sugar. The molar mass of carbon is 12.011 g/mol, hydrogen 1.008 g/mol, and oxygen 15.999 g/mol. Thus, the molar mass of table sugar will be:
[tex]12(12.011)+22(1.008)+11(15.999) = 342.297\text{ g/mol}[/tex]
To cancel units, we can write our ratio as:
[tex]\displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}[/tex]
With grams in the denominator.
And by definition:
[tex]\displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Combining the two ratios and the starting value, we acquire:
[tex]3420 \text{ g C$_{12}$H$_{22}$O$_{11}$}\cdot \displaystyle \frac{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}{342.297\text{ g C$_{12}$H$_{22}$O$_{11}$}}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1\text{ mol C$_{12}$H$_{22}$O$_{11}$}}[/tex]
Cancel like units:
[tex]=3420 \cdot \displaystyle \frac{1}{342.297}\cdot \displaystyle \frac{6.022\times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}}{1}[/tex]
Multiply:
[tex]\displaystyle = 60.1677... \times 10^{23} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
Rewrite:
[tex]\displaystyle = 6.01677... \times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}[/tex]
The resulting answer should have three significant digits. Hence:
[tex]=6.02\times 10^{24} \text{ C$_{12}$H$_{22}$O$_{11}$}}[/tex]
So, there are about 6.02 * 10²⁴ molecules of table sugar in 3.42 kilograms of the substance.
Answer:
2.69×10²³ molecules of PBr₅
6.02×10²⁴ molecules of C₁₂H₂₂O₁₁
Explanation:
To solve the first problem, we want to first find formula for phosphorus pentabromide, which is PBr₅. Now, we need to know the molar mass of PBr₅, which is about 430.49 g/mol. To get to molecules, we need to use Avogadro's number, which is 6.022×10²³ molecules/mol.
[tex]192g*\frac{1mol}{430.49g} *\frac{6.022*10^{23}molecules}{1mol} =2.69*10^{23} molecules[/tex]
Now, we know that there are about 2.69×10²³ molecules of PBr₅.
To solve the second problem, we need to use Avogadro's number, along with finding the molar mass of C₁₂H₂₂O₁₁, and converting kilograms to grams.
[tex]3.42 kg*\frac{1000g}{1kg} *\frac{1mol}{342.3g} *\frac{6.022*10^{23} molecules}{1mol} =6.02*10^{24} molecules[/tex]
Now, we know that there are about 6.02×10²⁴ molecules of C₁₂H₂₂O₁₁.
A solution is prepared at that is initially in diethylamine , a weak base with , and in diethylammonium chloride . Calculate the pH of the solution. Round your answer to decimal places.
Answer:
10.96
Explanation:
A solution is prepared at 25 °C that is initially 0.14 M in diethylamine, a weak base with Kb = 1.3 × 10⁻³, and 0.20 M in diethylammonium chloride. Calculate the pH of the solution. Round your answer to 2 decimal places.
Step 1: Calculate the pOH of the solution
Diethylamine is a weak base and diethylammonium (from diethylammonium chloride) its conjugate acid. Thus, they form a buffer system. We can calculate the pOH of this buffer system using the Henderson-Hasselbach's equation.
pOH = pKb + log [acid]/[base]
pOH = -log 1.3 × 10⁻³ + log 0.20 M/0.14 M
pOH = 3.04
Step 2: Calculate the pH of the solution
We will use the following expression.
pH + pOH = 14
pH = 14 - pOH = 14 -3.04 = 10.96
Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcohol with two other carbons attached to the carbon with the hydroxyl group_____.An alcohol with one other carbon attached to the carbon with the hydroxyl group____.An alcohol with three other carbons attached to the carbon with the hydroxyl group____.
Answer:
1). 1-pentanol - Primary
2). 3-ethyl-3-pentanol - Tertiary
3). 2-hexanol - Secondary
4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - Secondary
5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - Primary
6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - Tertiary
Explanation:
The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.
In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.
In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.
In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example: 3-ethyl-3-pentanol, -tert -butyl alcohol, etc.
The combustion of hydrogen-oxygen mixtures is used to produce very high temperatures (ca. 2500 °C) needed for certain types of welding operations. Consider the reaction to be
H₂(g) + ½O₂(g) → H₂O(g) ∆H° = -241.8 kJ/mol
What is the quantity of heat evolved, in kilojoules, when a 180 g mixture containing equal parts of H₂ and O₂ by mass is burned?
in kj
Answer:
1360kJ are evolved
Explanation:
When 1mole of H2 reacts with 1/2 moles O2 producing 1 mole of water and 241.8kJ.
To solve this question we need to find the limiting reactant knowing were added 90g of H2 and 90g of O2 as follows:
Moles H2 -Molar mass: 2g/mol-
90g H2 * (1mol / 2g) = 45 moles
Moles O2 -Molar mass: 32g/mol-
90g * (1mol / 32g) = 2.81moles
For a complete reaction of 2.81 moles of O2 are needed:
2.81 moles O2 * (1mol H2 / 1/2 mol O2) = 5.62 moles H2
As there are 45 moles, H2 is the excess reactant and O2 the limiting reactant.
As 1/2 moles O2 produce 241.8kJ, 2.81 moles will produce:
2.81 moles O2 * (241.8kJ / 1/2moles O2) =
1360kJ are evolvedThe quantity of heat evolved when 180 g mixture containing equal parts of H₂ and O₂ burned is
The equation for the combustion of hydrogen is given as:
[tex]\mathbf{H_2 + \dfrac{1}{2}O_2 \to H_2O \ \ \ \ \Delta H_r^0 = -241.8\ kJ/mol}[/tex]
Recall that:
number of moles = mass/molar mass:Since the mass of 180 g is equally shared by H₂ and O₂, then:
mass of H₂ = 90 gmass of O₂ = 90 gThe number of moles of the reactant can be determined as follows:
For H₂:
[tex]\mathbf{no \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{2.016 \ g/mol}}[/tex]
no of moles = 44.6 mol
For O₂:
[tex]\mathbf{no \ of \ moles = \dfrac{90 \ g }{32 \ g/mol}}[/tex]
no of moles = 2.8 mol
Here, since O₂ has a lesser amount of mole, then O₂ is regarded as the limiting reagent here:
If 1/2 moles of O₂ produces -241.8 kJ/mol of water;
Then, the quantity of heat that will evolve when 180 g mixture containing equal parts of H₂ and O₂ burned is:
[tex]\mathbf{= \Big (\dfrac{2.81 \ mol}{\dfrac{1}{2 } \ mol }\Big) \times (-241.8 \ kJ) }[/tex]
[tex]\mathbf{= \Big (5.62\Big) \times (-241.8 \ kJ) }[/tex]
= - 1358.91 kJ
≅ - 1360 kJ
Therefore, we can conclude that the quantity of heat evolved is - 1360 kJ
Learn more about the quantity of heat here:
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what is Lewis acid and Lewis base? give examples
Explanation:
example is copper iron...........
time is direct propotinally to rate of chemical reaction .explain if time is negretted and temperature remain costant?
Answer:
the constant temperature will be 12 .F bec in places it is cold
cho một thực phẩm có độ ẩm tương đối là 81 hỏi hoạt độ nước trong thực phẩm đó là bao nhiêu
Explanation:
For a food with a relative humidity of 81, what is the water activity in the food?
Water activity in a food = relative humidity / 100
Substitute the given values in this formula to get water activity:
Water activity in a food = 81 / 100
Hence, water activity =0.81
An example of a molecular compound that obeys the octet rule in which all atoms have a zero formal charge is:
Answer:
[tex]NCl_3[/tex]
Explanation:
An octet rule is a thumb rule in the chemical sciences in which there is a natural tendency for an atom to prefer eight electrons in the valence shell of the atom. When there are less than eight electrons in the atom, they react with other atoms and form more stable compounds.
In the context, nitrogen trichloride, [tex]NCl_3[/tex], is an example of molecular compound which obeys the octet rule having a zero formal charges on each atom of the compound.
Part A
If the theoretical yield of a reaction is 23.5 g and the actual yield is 14.8 g, what is the percent yield?
Answer:
[tex]\boxed {\boxed {\sf 63.0 \%}}[/tex]
Explanation:
The percent yield is the ratio of the actual yield to the theoretical yield.
[tex]percent \ yield = \frac{actual \ yield}{theoretical \ yield} * 100[/tex]
The actual yield is the amount obtained from performing a chemical reaction. For this problem, it is 14.8 grams. The theoretical yield is the potential amount from performing a chemical reaction at maximum performance. For this problem, it is 23.5 grams.We can substitute the known values into the formula.
[tex]percent \ yield= \frac{ 14.8 \ g}{23.5 \ g}*100[/tex]
Divide.
[tex]percent \ yield = 0.629787234043*100[/tex]
Multiply.
[tex]percent \ yield = 62.9787234043[/tex]
The original measurements for the theoretical and actual yields have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place.
The 7 to the right, in the hundredths place, tells us to round the 9 up to a 0. Since we rounded up to 0, we have to move to the next place to the left and round the 2 up to a 3.
[tex]percent \ yield \approx 63.0[/tex]
The percent yield is approximately 63.0 percent.
Help naming this plzzzzzzzzzzzzz
Answer:
A. 3-chloro-1-methylcyclobutane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).
Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.
Regards!
complete the following steps.
Remember to follow lower numbered rules first.
K2S(aq) + LiOH(aq)->KOH (?) + Li2S(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write “no
reaction.” (1 pt)
Answer:
See explanation
Explanation:
a) The balanced chemical reaction equation is;
2 LiOH + K2S ------> Li2S + 2 KOH
b) If we include the states of matter then we have;
2 LiOH(aq) + K2S(aq) -----> Li2S(s) + 2 KOH(aq)
c) Complete ionic equation;
2Li^+(aq) + 2OH^-(aq) + 2K^+(aq) + S^+(aq) ----> Li2S(s) + 2K^+(aq) + 2OH^-(aq)
Net ionic equation;
2Li^+(aq) + S^2-(aq) -----> Li2S(s)
This is the chemical formula for zinc bromate: . Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.
Answer:
30%
Explanation:
This is the chemical formula for zinc bromate: Zn(BrO₃)₂. Calculate the mass percent of oxygen in zinc bromate. Round your answer to the nearest percentage.
Step 1: Determine the mass of 1 mole of Zn(BrO₃)₂
M(Zn(BrO₃)₂) = 1 × M(Zn) + 2 × M(Br) + 6 × M(O)
M(Zn(BrO₃)₂) = 1 × 65.38 g/mol + 2 × 79.90 g/mol + 6 × 16.00 g/mol
M(Zn(BrO₃)₂) = 321.18 g/mol
Step 2: Determine the mass of oxygen in 1 mole of Zn(BrO₃)₂
There are 6 moles of atoms of oxygen in 1 mole of Zn(BrO₃)₂.
6 × m(O) = 6 × 16.00 g = 96.00 g
Step 3: Calculate the mass percent of oxygen in Zn(BrO₃)₂
%O = mO/mZn(BrO₃)₂ × 100%
%O = 96.00 g/321.18 g × 100% ≈ 30%
Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is ________ M.
A) 0.0735
B) 0.0762
C) 0.0980
D) 0.0709
E) 0.00253
Answer: The concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.
Explanation:
Given: Concentration of hydrogen fluoride = 0.126 M
Concentration of fluoride ions = 0.1 M
Volume of HCl = 9.0 mL
Concentration of HCl = 0.01 M
Volume of HCl = 25.0 mL
Moles of [tex]F^{-}[/tex] ions are calculated as follows.
[tex]Moles of F^{-} = molarity \times volume\\= 0.1 M \times 0.025 L\\= 0.0025 mol[/tex]
Moles of HF are as follows.
[tex]Moles of HF = Molarity \times Volume\\= 0.126 M \times 0.025 L\\= 0.00315 mol[/tex]
Moles of HCl are as follows.
[tex]Moles of HCl = Molarity \times volume\\= 0.01 M \times 0.009 L\\= 0.00009 mol[/tex]
Now, reaction equation with initial and final moles will be as follows.
[tex]H^{+} + F^{-} \rightarrow HF[/tex]
Initial: 0.00009 0.0025 0.00315
Equilibrium: (0.0025 - 0.00009) (0.00315 + 0.00009)
= 0.00241 = 0.00324
Total volume = (9.00 mL + 25.0 mL) = 34.0 mL = 0.034 L
Hence, concentration of fluoride ions is calculated as follows.
[tex]Concentration = \frac{moles}{volume}\\= \frac{0.00241 mol}{0.034 L}\\= 0.0709 M[/tex]
Thus, we can conclude that concentration of fluoride ions after the addition of 9.00 mL of 0.0100 M HCl to 25.0 mL of this solution is 0.0709 M.
write down any two test for the CH2=CH2[ethene]
Answer:
a) Linear polymerization
b) cyclic polymerization
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
3) Convert 0.250 moles of aluminum sulfate to grams.
4) Convert 2.70 grams of ammonia to moles.
Answer:
0.000731 grams aluminium sulfate
46.0 mols ammonia
Explanation:
ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol
[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]
NH3 has a molar mass of 17.031 g/mol
[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]
Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]
we have to find the 0.250 moles of aluminum sulphate.
[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]
[tex]\\\\\\[/tex]
Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]
We have to find 2.70 grams of ammonia
[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]
Which of the following is not organic compound?
a. CH4
b. H2CO3
c. CCl4
d. CH3-OH
For a given fluorophore, select the choice that correctly lists the processes of fluorescence, absorption, and phosphorescence in order from shortest to longest wavelength.
a. absorption, fluorescence, phosphorescence
b. Fluorescence = phosphorescence, absorption
c. fluorescence, phosphorescence, absorption
d. phosphorescence, fluorescence, absorption
e. absorption, phosphorescence, fluorescence
f. absorption, fluorescence = phosphorescence
Answer:
absorption, fluorescence = phosphorescence
Explanation:
Given a particular fluorophore, the wavelength of absorption of the fluorophore is always shorter. Both fluorescence and phosphorescence are kinds of photoluminescence.
Recall that both fluorescence and phosphorescence occur at a longer wavelength. The difference between the two is only in the time taken during the process. While fluorescence takes a shorter time to occur, phosphorescence takes a longer time to occur.
The major difference between fluorescence and phosphorescence is that change of spin occurs during phosphorescence but not fluorescence.
When you burn a birthday candle, you may wonder whether the fire you see is matter. The
flame consists of hot, glowing gases, such as oxygen, carbon dioxide, water vapor, and
parts of the candle that have been vaporized. The heat and light given off are forms of
energy. The smoke contains ash and unburned particles.
a. From the above description of a candle flame, list at least three things that are matter and
three things that are not mattel. (6 points)
Matter
Not Matter
Answer:
matter
1. Candle. not matter
1. light
2. Unburned Particles
2. heat
3. Ash
3. energy
Explanation:
1.The candle is experiencing a solid phase into a gas phase because the heat given off causes smoke.
2. Chemical Potential Energy to Heat EnergyThe candle has Chemical Potential Energy then when it gets lit by the flame heat energy is released.
The smallest atoms can themselves exhibit quantum-mechanical behavior. Calculate the de Broglie wavelength (in picometers) of a hydrogen atom traveling at 490 m/s. Express your answer using three significant figures.
Answer:
810 pm
Explanation:
Step 1: Given and required data
Velocity of the atom (v): 490 m/sMass of a hydrogen atom (m): 1.67 × 10⁻²⁷ kgPlanck's constant (h): 6.63 × 10⁻³⁴ J.sStep 2: Calculate the de Broglie wavelength of the hydrogen atom
We will use de Broglie's equation.
λ = h / m × v
λ = 6.63 × 10⁻³⁴ J.s / 1.67 × 10⁻²⁷ kg × 490 m/s = 8.10 × 10⁻¹⁰ m
Step 3: Convert 8.10 × 10⁻¹⁰ m to picometers
We will use the conversion factor 1 m = 10¹² pm.
8.10 × 10⁻¹⁰ m × 10¹² pm/1 m = 810 pm
How many molecules of Iron(II)oxide are present in 35.2*10^-23 g of Iron (II)oxide?
Answer:
R.F.M of Iron (II) oxide :
[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]
Moles :
[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]
Molecules :
[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]
The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.
What is Avogadro's number?Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.
The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.
Given, the mass of the iron oxide = 35.2 ×10⁻²³ g
The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol
71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³
35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)
The number of molecules of FeO in a given mass = 2.95 molecules
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What type of reaction occurs? S(s) + O2 (g) → SO2(g)
Answer:
reductants or redox reactions
Curium – 245 is an alpha emitter. Write the equation for the nuclear reaction and identify the product nucleus.
Answer:
Please find the complete solution in the attached file.
Explanation:
Question 2 10
10 Points
Which of the following chemical equations depicts a balanced chemical equation?
O A. AgNO, Kcro > KNO, Agro,
OB. AgNO3 + Kycro, » 2K NO; + Agro,
C.3AgNO3 + 2K,Cro--> 3KNO3 + 249900,
D. 2AgNO, K Cro-> 2KNO; 1Cro,
Resol Selection
Answer:
2AgNO, K Cro-> 2KNO; 1Cro,
When do epa’s refrigerant management regulations exempt refrigerant from the venting prohibition
The Environmental Protection Agency (EPA) refrigerant management regulations would exempt any refrigerant from venting problem when it determines that the refrigerant in an appliance do not pose a threat to the environment (surrounding) if released.
A refrigerant can be defined as any chemical substance that undergoes a phase change (liquid and gas) so as to enable the cooling and freezing of materials. They are typically used in air conditioners, refrigerators, water dispensers, etc.
In the United States of America, the Environmental Protection Agency (EPA) is a governmental agency which was established by U.S Congress and it is saddled with the responsibility of overseeing all aspects of pollution, environmental clean up, degradation, pesticide use, contamination, and hazardous waste spills. Also, EPA research solutions, policy development, and enforcement of regulations through the resource Conservation and Recovery Act.
Simply stated, the Environmental Protection Agency (EPA) is the governmental agency set up to ensure that various industries, factories and people comply with laws and regulations concerning the environment.
In conclusion, appliances such as a refrigerator, air conditioner (AC), etc., whose refrigerants do not pose a threat (potentially cause damage) to the environment if released are typically exempted by the EPA's refrigerant management regulations.
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300g de acido comercial se disuelve en agua destilada contenidos en un cono, cuyo radio es de 0.005Km y 300cm de altura, si la densidades de 1.2g/m3 ¿Cuál es la concentración expresada en %m/m?
Answer:
97.0%m/m es la concentración de la solución
Explanation:
El porcentaje masa/masa (%m/m) se define como 100 veces el radio entre la masa de soluto (300g de HCl) y la masa de la solución. Para hallar la masa de la solución debemos hallar la masa del agua (Solvente) haciendo uso de la ecuación del volumen de un cono. Con el volumen del cono podemos hallar la masa del agua haciendo uso de la densidad, así:
Volumen:
Volumen Cono = π*r²*h / 3
Donde r es el radio = 0.300m
h la altura = 5m
Volumen = π*(5m)²*0.300m / 3
7.85m³
Masa Agua:
7.85m³ * (1.2g / m³) = 9.42g Agua
Masa solución:
300g HCl + 9.42g Agua = 309.42g Solución
%m/m:
300g HCl / 309.42g * 100 =
97.0%m/m es la concentración de la soluciónHi,Valency of chlorine is 1. Why?Thank you
Answer:
The chlorine element belongs to group 17 because it has 7 valence electron . Its valency is 1 . It can gain one electron from any other atom to become stable. This means that it can never form a double or triple bond.
