Answer:
Fe3(PO4)2
Explanation:
calculate the molar mass for each of the following compounds 7. PbSO
8. Ca(OH)2
9. Na3PO4
10. (NH4)2CO3
11. C6H12O6
12. Fe3(PO4)2
13. (NH4)2S
14. Zn(C2H3O2)2
what is valency of an atom?
The number of replaceable electrons in an atom is called its valency.
Examples
Monovalent - HydrogenDivalent - OxygenValency = 8 - Number of electron in last shell [When number of electrons in last shell > 4]Valency = Number of electron in last shell [When number of electrons in last shell < 4]Thanks !
☺️☺️☺️☺️☺️☺️☺️
Answer:
the combining capacity if an atom is know as valency.
the property of an element that determines the number of other atimd with an aton if the element can combine.
Identify what reagents you would use to achieve each transformation: Conversion of 2-methyl-2-butene into a secondary alkyl halide. Br2, ROOR Br2, H2O HBr, ROOR HBr Conversion of 2-methyl-2-butene into a tertiary alkyl halide. Br2, H2O HBr Br2, ROOR HBr, ROOR
Answer:
Conversion of 2-methyl-2-butene into a secondary alkyl halide - ROOR, HBr
Conversion of 2-methyl-2-butene into a tertiary alkyl halide - HBr
Explanation:
The addition of HBr to 2-methyl-2-butene occurs in accordance to Markovnikov rule in the absence of peroxide.
According to Markovnikov rule; ''the negative part of the addendum is attached to the carbon atom bearing the least number of hydrogen atoms.'' Following the Markovnikov rule, the tertiary alkyl halide is obtained.
In the presence of peroxide, this rule is not followed and the reaction proceeds in an anti-Markovnikov way to yield a secondary alkyl halide.
. Which of the following statement is not related to a chemical reaction ? A. New substances are formed B. Atoms of the elements transform into atoms of other elements C. The properties of the new substances will be different D. There will be bond breaking and bond forming
Answer:
the answer should be B because elements do not tranform into other elements in a chemical reaction
am I right please?
When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are absorbed. CO(g) + H2O(l)CO2(g) + H2(g) Is this reaction endothermic or exothermic? _________ What is the value of q? kJ
You have been contracted to determine how different salts affect the pH of water. Which of the solids in the following set should you test to investigate for the effects of cations on pH?
a. AlBr3
b. Rb2SO3
c. MgCl2
d. RbBrO
e. CH3NH3Br
Answer:
Hence the solids that should test to investigate the effects of cations on pH is
[tex]AlBr_{3}[/tex] (Cation is Al 3+)
[tex]MgCl_{2}[/tex] ( Cation is Mg 2+)
[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).
Explanation:
The solids in the following should you test to investigate the effects of cations on pH.
[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)
[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)
[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )
The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.
What are cations and pH?Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.
The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.
Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.
Therefore, absorption of the cation reduces the pH.
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Which subshells are found in each of the following shells
electron subshell - M shell
Answer:
3
Explanation:
The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.
As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.
Hence, M shell contains s,p and d subshells.
The molecular ion is not visible in the mass spectrum of 2-chloro-2- methylpropane. At what m/z value would the molecular ion be if it were visible? What evidence is there in the mass spectrum that suggests that the peak at m/z= 77 contains a chlorine atom?
Answer: hello the complete question is attached below
Visibility of molecular ion = m/z value of 77
Explanation:
For The molecular ion to be visible, it has to be at an m/z value of 77 and this is because molecular ions will have an m/z ratio = molecular mass of given molecule in most cases but not always in all cases.
And the visibility is possible after the removal of CH₃ ion.
ii) Evidence in the mass spectrum that suggests peak at m/z = 77
attached below
Which is a statement of cell theory? All cells are made up of living molecules. All plants are made of cells. All animals are made of cells. All cells are produced from other cells.
Answer:
all cells are produced from other preexisting cells through cell division
If 7 mol of copper reacts with 4 mol of oxygen, what amount of copper (II) oxide is produced? What amount of the excess reactant remains?
Answer:
7 mol CuO
0.5 mol O₂
Explanation:
Step 1: Write the balanced equation
2 Cu + O₂ ⇒ 2 CuO
Step 2: Identify the limiting reactant
The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.
The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.
Since TMR > EMR, Cu is the limiting reactant
Step 3: Calculate the amount of CuO produced
7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO
Step 4: Calculate the excess of O₂ that remains
The amount of O₂ that reacts is:
7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂
The excess of O₂ that remains is:
4 mol - 3.5 mol = 0.5 mol
In an analysis of interhalogen reactivity, 0.350 mol ICl was placed in a 5.00 L flask and allowed to decompose at a high temperature.
2 ICl(g) I2(g) + Cl2(g)
Calculate the equilibrium concentrations of I2, Cl2, and ICl. (Kc = 0.110 at this temperature.)
I2 M
Cl2 M
ICl M
Answer:
[ICl] = 0.0420 M
[I₂] = [Cl₂] = 0.0140 M
Explanation:
Step 1: Calculate the initial concentration of ICl
[ICl] = 0.350 mol / 5.00 L = 0.0700 M
Step 2: Make an ICE chart
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
I 0.0700 0 0
C -2x +x +x
E 0.0700-2x x x
The concentration equilibrium constant (Kc) is:
Kc = 0.110 = [I₂] [Cl₂] / [ICl]² = x² / (0.0700-2x)² = (x/0.0700-2x)²
0.332 = x/0.0700-2x
x = 0.0140
The concentrations at equilbrium are:
[ICl] = 0.0700-2x = 0.0700-0.0280 = 0.0420 M
[I₂] = [Cl₂] = x = 0.0140 M
The enthalpy of vaporization of water is 2,257,000 J/kg. If I have a 1 kg sample, how much energy is needed to boil all of it
Answer:
2257000 J
Explanation:
Applying,
Q = Cₓm.................. Equation 1
Where Q = amount of energy need to boil the water, Cₓ = Enthalpy of vaporization of water, m = mass of water.
From the question,
Given: Cₓ = 2257000 J/kg, m = 1 kg
Substitute these values into equation 1
Q = 2257000×1
Q = 2257000 J
Hence the energy needed to boil all of the water is 2257000 J
Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?
a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood
Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.
Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.
A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.
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A metal (C = 0.2158 cal/g· °C) is removed from a hot (350. °F) oven in which it had achieved thermal equilibrium. The metal is placed into 200. mL acetic acid. The temperature of the acid increases to 90.3 °C from 24.3 °C. What is the mass of the metal? (dacetic acid = 1.04 g/cm3; Cs, acetic acid = 2.055 J/g·°C) Group of answer choices 120. g 362 g 1452 g 347 g 281 g
Answer:
362g
Explanation:
heat lost by metal= heat gained by acetic acid
tfs are the same so you cando delta T
convert Cal/gc to J/gc
thectgod ig follow
Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 hours for an adolescent. What fraction of the original reactant concentration will remain after 8 hours if the original concentration was 8.4 x 10-5 M.
Explanation:
The given data is:
The half-life of gentamicin is 1.5 hrs.
The reaction follows first-order kinetics.
The initial concentration of the reactants is 8.4 x 10-5 M.
The concentration of reactant after 8 hrs can be calculated as shown below:
The formula of the half-life of the first-order reaction is:
[tex]k=\frac{0.693}{t_1_/_2}[/tex]
Where k = rate constant
t1/2=half-life
So, the rate constant k value is:
[tex]k=\frac{0.693}{1.5 hrs}[/tex]
The expression for the rate constant is :
[tex]k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}[/tex]
Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:
[tex]\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6[/tex]
Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
What minimum mass of HCl in grams would you need to dissolve a 2.2 g iron bar on a
padlock?
2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.
What is dissolution?When a solute is dissolved in a solvent, a solution is created. Dissolution is the process through which solutes, or dissolved parts, combine to form a solution inside a solvent. In this procedure, the gas, liquid, or solid dissolves inside the original solvent and forms a solution.
In some polymer applications, dissolution is also an issue since it results in swelling, a loss of strength and stiffness, and a change in volume. Whether a chemical process is man-made or natural, dissolution is crucial. Catalysts are tested using dissolution. 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.
Therefore, 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.
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Can someone teach me step by step how finding the oxidation number in this problem:
Fe in Fe(CIO2)3
Answer:
+3
Explanation:
u see sum of oxidation number in all situations have to be 0
ClO2 =-1
so Fe is +3
Buffer solutions that maintain certain levels of pH or acidity are widely used in biochemical experiments. One common buffer system uses sodium dihydrogenphosphate and sodium monohydrogenphosphate. What are the formulas of these two compounds
Answer:
Sodium dihydrogenphosphate = NaH₂PO₄
Sodium monohydrogenphosphate = Na₂HPO₄
Explanation:
A buffer solution is a solution is a solution that resists changes to its oH when a little quantity of strong acid or strong base is added to it.
They are solutions of weak acids or weak bases and their salts known as conjugate base or conjugate acids respectively for the weak acids and weak bases.
For example, a solution of the weak acid ethanoic acid and its salt or conjugate base, sodium ethanoate serves as a buffer solution.
In biochemical experiments, where the pH of the reaction medium is kept as constant and as close as possible to that of the internal environment, buffer solutions are widely used. One of the commonly used buffers is the phosphate buffer. The phosphate buffer consists of the acid salts sodium dihydrogenphosphate and sodium monohydrogenphosphate. Sodium dihydrogenphosphate serves as the weak acid while sodium monohydrogenphosphate serves as the conjugate base.
The formulas of these two compounds are given below:
Sodium dihydrogenphosphate = NaH₂PO₄
Sodium monohydrogenphosphate = Na₂HPO₄
What are the effects of global warming?
the effects are: temperature rises, water shortages, and increased fire threats
Given the following balanced equation:
3Cu(s) + 8HNO3(aq) = 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
Determine the mass of copper (II) nitrate that would be formed from the complete reaction
of 35.5g of copper with an excess of nitric acid.
Answer: The mass of copper (II) nitrate produced is 105.04 g.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of copper = 35.5 g
Molar mass of copper = 63.5 g/mol
Plugging values in equation 1:
[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]
The given chemical equation follows:
[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]
By the stoichiometry of the reaction:
If 3 moles of copper produces 3 moles of copper (II) nitrate
So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate
Molar mass of copper (II) nitrate = 187.56 g/mol
Plugging values in equation 1:
[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]
Hence, the mass of copper (II) nitrate produced is 105.04 g.
Forcus on the yellow highlighted texts, your help is appreciated.
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Answer:
Transition temperature is the temperature at which a substance changes from one state to another.
Allotropy is the existence of an element in many forms.
The density of a gas cannot be measured.
True
False
Answer:
False
Explanation:
Answer:
False
Explanation:
How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution
Answer:
89.04 g of NaNO₃.
Explanation:
We'll begin by converting 838 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
838 mL = 838 mL × 1 L / 1000 mL
838 mL = 0.838 L
Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:
Volume = 0.838 L
Molarity = 1.25 M
Mole of NaNO₃ =?
Mole = Molarity × volume
Mole of NaNO₃ = 1.25 × 0.838
Mole of NaNO₃ = 1.0475 mole
Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:
Mole of NaNO₃ = 1.0475 mole
Molar mass of NaNO₃ = 23 + 14 + (16×3)
= 23 + 14 + 48
= 85 g/mol
Mass of NaNO₃ =?
Mass = mole × molar mass
Mass of NaNO₃ = 1.0475 × 85
Mass of NaNO₃ = 89.04 g
Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.
Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 space O subscript 2 space (g )space rightwards arrow 3 space C O subscript 2 space (g )space plus space 4 space H subscript 2 O space (g )How many liters of carbon dioxide are produced at STP when 44 g of C3H8 completely reacts with oxygen
Explanation:
The balanced chemical equation of the reaction is:
[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]
From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
explain why it is important not to correct any gas from the first few seconds of the experiment
Answer:
gu kha fuschhehdjdvdbeodbr
pls help name any of these compounds
Answer:
D. Propanol
Explanation:
C3H7OH the presence of alcohol functional group makes it propanol
What is alkaline and what is acidic pH
Answer:
An alkaline is a substance that dissolves in water to produce hydroxyl ions (OH-)
Explanation:
The pH range of an alkaline is from 8–14.
Acidic pH ranges from 0–6.9.
Complete the balanced dissociation equation for the compound below in aqueous solution. If the compound does not dissociate, write NR after the reaction arrow.
HI (aq) -->
Answer:
[tex]{ \bf{HI _{(aq)} \: → \: H {}^{ + } _{(aq)} \: + \: \: I {}^{ - } _{(aq)} }}[/tex]
What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with
4.00g of hydrogen?
P4(g)+ 6H2(g) → 4PH3(g)
Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g
Explanation:
Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.
Excess reagent is defined as the reagent which is left behind after the completion of the reaction.
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For [tex]P_4[/tex]:Given mass of [tex]P_4[/tex] = 62.0 g
Molar mass of [tex]P_4[/tex] = 124 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]
For [tex]H_2[/tex]:Given mass of [tex]H_2[/tex] = 4.00 g
Molar mass of [tex]H_2[/tex] = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]
The chemical equation follows:
[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]
By stoichiometry of the reaction:
If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]
So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]
As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]
So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]
We know, molar mass of [tex]PH_3[/tex] = 34 g/mol
Putting values in equation 1, we get:
[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]
Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g
Calculate the mass of water produced when 7.49 g of butane reacts with excess oxygen.
Answer:
[tex]m_{H_2O}=12.9gH_2O[/tex]
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the reaction whereby butane is combusted in the presence of excess oxygen:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
Thus, we can evidence a 2:10 mole ratio of butane to water, and thus, the stoichiometric setup to calculate the mass of produced water is:
[tex]m_{H_2O}=7.49gC_4H_{10}*\frac{1molC_4H_{10}}{52.12gC_4H_{10}} *\frac{10molH_2O}{2molC_4H_{10}}*\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=12.9gH_2O[/tex]
Regards!
