Answer:
5.00 mol Mg
10.0 mol Cl
40.0 mol O
Explanation:
Step 1: Given data
Moles of Mg(ClO₄)₂: 5.00 mol
Step 2: Calculate the number of moles of Mg
The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.
5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg
Step 3: Calculate the number of moles of Cl
The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.
5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl
Step 4: Calculate the number of moles of O
The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.
5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O
how many moles of ammonia can be formed from 4.0 mol H2
how many moles of hydrogen are required to form 8.0 mol NH3
how many moles of nitrogen are required to react with 6.0 mol H2
Answer: 1. 2.7 moles of ammonia are formed
2. 12.0 moles of hydrogen are required
3. 2.0 moles of nitrogen are required
Explanation:
The balanced chemical equation is:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
According to stoichiometry:
3 moles of hydrogen form = 2 moles of ammonia
Thus 4.0 moles of hydrogen form =[tex]\frac{2}{3}\times 4.0=2.7moles[/tex] of ammonia
According to stoichiometry:
2 moles of ammonia are formed by = 3 moles of hydrogen
Thus 8.0 moles of ammonia are formed by =[tex]\frac{3}{2}\times 8.0=12.0moles[/tex] of hydrogen
According to stoichiometry:
3 moles of hydrogen react with = 1 mole of nitrogen
Thus 6.0 moles of hydrogen react with =[tex]\frac{1}{3}\times 6.0=2.0moles[/tex] of nitrogen
How many grams (g) are in 4.00 moles of Carbon Dioxide (CO2) ?
Answer:
m = 176.04 g .
Explanation:
Hello there!
In this case, according to the mole-mass relationships, which are based off the mass of one mole of any compound via the molar mass, it is possible to realize that the molar mass of carbon dioxide is 44.01 (12.01+16*2) g/mol, and therefore, the mass in grams of 4.00 moles of this compound are calculated as shown below:
[tex]m=4.00molCO_2*\frac{44.01gCO_2}{1molCO_2}\\\\m=176.04g[/tex]
Best regards!
Plz help me I am timed!!!!!
Answer:
i think its the gibbous phase
Explanation:
How many moles are in 272 grams of hydrogen peroxide (H2O2) ?
Answer:
8 moles
Explanation:
When we are asked to convert from grams of a substance into moles, we have to use the substance's molar mass.
Meaning that for this problem, we'll use the molar mass of hydrogen peroxide (H₂O₂), as follows:
272 g ÷ 34 g/mol = 8 molThere are 8 moles in 272 grams of hydrogen peroxide.
Paul swam 7 5/8 miles. His sister swam five times as many miles. How many miles did Paul's sister swim?
Answer:
Distance cover by Paul's sister by swim = 38.125 miles
Explanation:
Given:
Distance cover by Paul by swim = [tex]7\frac{5}{8}[/tex] = 61/8 miles
Distance cover by Paul's sister by swim = 5 times Distance cover by Paul
Find:
Distance cover by Paul's sister by swim
Computation:
Distance cover by Paul's sister by swim = 5 times Distance cover by Paul
Distance cover by Paul's sister by swim = 5 x [61 / 8 miles]
Distance cover by Paul's sister by swim = 305 / 8 miles
Distance cover by Paul's sister by swim = 38.125 miles
Acidic solutions have pH value less than 7 . Select one : O True O False
Answer:
Explanation:
The answer is true.
Answer:
true
Explanation:
acids have ph which is less than 7 and base have ph greater than 7
Calculate the pH of a 0.010 M NaNO2 solution.
Answer:
Calculate the pH of 0.010 M HNO2 solution. The K, for HNO2 is 4.6 x 104
Answer: pH = 2.72
Calculate the pH of each solution:
0.030 M HBr
3.0 M HF
0.50 M HNO3
O.0010 M NaOH
0.150 M KOH
When you're just chilling in the pool and then you think about this:
Explanation:
Sonic is my bf forever, back off
Answer:
oof I always lose on video games so I just quit
Which statement describes a homogeneous catalyst?
do
It is in a gaseous phase only. It is in the same phase as the reactants.
It is completely consumed by reactants.
It forms different products than the uncatalyzed reaction forms.
The statement which is given below describe a homogeneous catalyst.
"It is in the same phase as the reactants"
What is homogeneous catalyst?The catalyst which have similar phase to the reactant of the reaction is called homogeneous catalyst.
Mostly in homogeneous catalyst, catalyst present in liquid form.
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Ive finished my class good luck!!? ❤️✌️
Answer:
yayyy! proud of u
Explanation:
The tablets were crushed, and 4.9993 g of the powder was transferred to a beaker and reacted with HCl. After filtration, the filtrate was transferred to a 100-mL volumetric flask and diluted with water. 20.00 mL of this stock solution were combined with 0.2 M Na3PO4. The resulting precipitate weighed 0.3451 g after drying. Calculate the moles of BiPO4 precipitated, the moles of Bi3 in the stock solution, and the mass of BSS per tablet.
Answer:
Explanation:
From the information given:
Mass of BiPO₄ = 0.3451 g
Number of moles of BiPO₄ = [tex]0.3451 \ g \ BiPO_4 \times \dfrac{1 \ mol \ BiPO_4}{303.95 \ g \ BiPO_4}[/tex]
[tex]= 0.001135 \ mol[/tex]
The number of moles of Bi³⁺ in 20.00 mL is:[tex]= 0.001135 \ mol \ BiPO_4 \times \dfrac{1 \ mol \ of \ Bi^{3+}}{1 \ mol \ BiPO_4}[/tex]
= 0.001135 mol of Bi³⁺
The number of moles of Bi³⁺ in 100 mL stock solution
[tex]= 0.001135 \ mol \ Bi^{3+} \times \dfrac{100 \ mL}{20.0 \ mL}[/tex]
[tex]= 0.005675 \ mol[/tex]
Mass of BSS in 4.9993 g tablets
[tex]m = 0.005675 \ mol \ Bi^{3+} \times \dfrac{1 \ mol \ BSS}{1 \ mol \Bi^{3+}} \times \dfrac{362.1 \ g \ BSS}{1 \ mol \ BSS}[/tex]
m = 2.055 g BSS
Mass of BSS in 5.0103 g (5 tables)
[tex]m = 2.055 g \ BSS \times \dfrac{5.0103 \ g}{4.9993 \ g}[/tex]
= 2.06 g
∴
The mass of BSS per tablet is [tex]=\dfrac{2.06 \ g}{5 \ tablet}[/tex]
= 0.412 g BSS/ tablet
Anyone know the answer
Answer:
b
Explanation:
i know
D
Question 2
What is the purpose of looking at chemical reactions?
to classify the type of reaction it is
to identify the type of reaction it is
to see how elements rearrange and represent something else
all of the above
Answer:
all of the above
Explanation
Plzzz help me plz I am timed!!
Answer:
Explanation:
pretty sure its gaillieo
Answer:
Galileo
Explanation:
he was the first one to use a telescope
Which expression represents the concentration of OH– in solution?
a. 10–14 / [H3O+]
b. [OH–] / 10–14
c. 10–14 – [H3O+]
d. 10–14 x [H3O+]
Answer:
c. 10–14[H3O+]
Explanation:
On a calculator, calculate 10-8.34, or "inverse" log ( - 8.34). Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10-5 M? The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH.
A low pitch sound is associated with which of the following?
а. high frequency
b. low frequency
с. fundamental tone
d. low intensity
No links plz
1. What do producers do, and what is another name for them?
Producers are the foundation of every food web in every ecosystem—they occupy what is called the first tropic level of the food web. The second trophic level consists of primary consumers—the herbivores, or animals that eat plants. At the top level are secondary consumers—the carnivores and omnivores who eat the primary consumers. Ultimately, decomposers break down dead organisms, returning vital nutrients to the soil, and restarting the cycle. Another name for producers is autotrophs, which means “self-nourishers.” There are two kinds of autotrophs. The most common are photoautotrophs—producers that carry out photosynthesis. Trees, grasses, and shrubs are the most important terrestrial photoautotrophs. In most aquatic ecosystems, including lakes and oceans, algae are the most important photoautotrophs.
Determine the mass of hydrogen contained in 9.06 x 1024 H2O molecules.
Answer:
15.2 g H2
Explanation:
2H2O -> 2H2 + O2
9.06 x 10^24 molecules x (1 mol/6.022 x 10^23 molecules) x (2 mol H2/2 mol H2O) x (1.008 g/1 mol) = 15.2 g H2
As a future educator how will you handle a pupil who could possibly be a gifted child
Answer:
Learn how gifted students think. ...
Created tiered assignments for students. ...
Include a variety of levels in your classroom library. ...
Utilize their talents and interests. ...
Explore real-word application.
When planning a simple experiment, what does it mean to “test one variable”?
a During the experiment, the scientist keeps the control the same but changes many other elements he or she is testing.
b During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis.
c The scientist can only use simple language and materials when planning the one variable to be tested.
d The scientist plans and performs just one science experiment a day to test one variable.
Answer:
I believe the answer is "b". "During the experiment, the scientist has only one element, or variable, that is changed to test the hypothesis."
Explanation:
I remember from last year but I'm not totally sure. Good luck!
The experimental setup has been the design that includes the experimental and the control group. during the experiment, only one variable is changed to test the hypothesis. Thus, option b is correct.
What is experimental design?The experiment design includes the testing of the hypothesis to solve the problem and reach the solution through the analysis of the observations. It has been the setup that includes the test and the control group that involves the dependent, independent, and controlled variables.
Testing one variable in the experimental designs means changing one of the elements of the experience to prove the reliability of the hypothesis. This variable is the factor that alters and affects the other factors of the experiment and leads to a cause and effect.
The result of the effect of the hypothesis proves the testability of the experiment and provides solutions to the research question. This further leads to the formulation of the law and theory.
Therefore, option A. testing one variable involves proving the hypothesis.
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Lead (ll) iodide (PbI2) has a solubility of 1.52×10 to the -3 mol/L.
1. write the dissolution reaction to PbI2 including all states.
2. Write the expression for Ksp for Pbl2.
3. What is the concentration of Pb2+ in the equilibrium solution?
4. What is the concentration of I- in the equilibrium solution?
5. Calculate the solubility product of Pbl2.
Answer:
A. PbI2(s) ===> Pb2+(aq) + 2I-(aq)
B. Ksp = [Pb2+][I-]^2
C. 1.52 x 10^-3 M. It is equal to the moles/L of PbI2 that go into solution.
D. 2 x 1.52 x 10^-3 = 3.04 x 10^-3 M
E. Ksp = (1.52x10^-3)(2.31x10^-6) = 3.51 x 10^-9
Explanation:
In the given question, [tex]\rm PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)[/tex] is the dissolution reaction of [tex]\rm PbI_2[/tex] in water, [tex]\rm Ksp = [Pb^{2+}][I^-]^2[/tex] is expression for Ksp for [tex]\rm PbI_2[/tex], [tex]1.52\times 10^{-3 }[/tex] mol/L is the concentration of [tex]\rm Pb^{2+ }[/tex] in the equilibrium solution, [tex]3.04\times 10^{-3}[/tex] mol/L is the concentration of I- in the equilibrium solution and
[tex]1.40\times 10^{-8}[/tex] is the solubility product of [tex]\rm PbI_2[/tex], respectively.
A reaction is a process that involves the transformation of one or more substances into one or more different substances.
1. The dissolution reaction of [tex]\rm PbI_2[/tex] in water is:
[tex]\rm PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)[/tex]
2. The expression for Ksp for [tex]\rm PbI_2[/tex] is:
[tex]\rm Ksp = [Pb^{2+}][I^-]^2[/tex]
Where, [tex]\rm [Pb^{2+}][/tex] is the concentration of [tex]\rm Pb^{2+ }[/tex] ions in solution and [tex]\rm [I^-][/tex] is the concentration of [tex]\rm I^-[/tex] ions in solution.
3. The solubility of [tex]\rm PbI_2[/tex] is [tex]1.52\times 10^{-3 }[/tex] mol/L. Since [tex]\rm PbI_2[/tex] dissociates into one [tex]\rm Pb^{2+ }[/tex] ion and two [tex]\rm I^-[/tex] ions, the concentration of [tex]\rm Pb^{2+ }[/tex] in solution is equal to the solubility of [tex]\rm PbI_2[/tex], which is [tex]1.52\times 10^{-3 }[/tex] mol/L.
[tex]\rm [Pb^{2+}] = 1.52\times 10^{-3}\ mol/L[/tex]
4. Since [tex]\rm PbI_2[/tex] dissociates into one [tex]\rm Pb^{2+ }[/tex] ion and two [tex]\rm I^-[/tex] ions, the concentration of [tex]\rm I^-[/tex] ions in solution is twice the solubility of [tex]\rm PbI_2[/tex] .
[tex]\rm [I^-] = 2 \times 1.52\times 10^{-3 }\ mol/L[/tex]
= [tex]3.04\times 10^{-3}[/tex] mol/L
5. The solubility product of [tex]\rm PbI_2[/tex] can be calculated using the expression for Ksp and the concentrations of [tex]\rm Pb^{2+ }[/tex] and [tex]\rm I^-[/tex] ions in solution.
[tex]\rm Ksp = [Pb^{2+}][I^-]^2[/tex]
= [tex]1.52\times 10^{-3 }[/tex] [tex]\times[/tex] [tex]\rm (3.04\times 10^{-3} mol/L)^2[/tex]
= [tex]1.40\times 10^{-8}[/tex]
Therefore, the dissolution reaction of [tex]\rm PbI_2[/tex] in water, expression for Ksp for [tex]\rm PbI_2[/tex], the concentration of Pb2+ in the equilibrium solution, the concentration of I- in the equilibrium solution and solubility product of [tex]\rm PbI_2[/tex] is mentioned above.
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Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Determine the limiting reactant in a mixture containing 169 g of RuI3, 58.0 g of CO, and 96.2 g of Ag. Calculate the maximum mass (in grams) of ruthenium pentacarbonyl, Ru(CO)5, that can be produced in the reaction. The limiting reactant is:
Answer:
71.6 g of Ru(CO)₅ is the maximum mass that can be formed.
The limiting reactant is Ag
Explanation:
The reaction is:
RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)
Firstly we determine the moles of each reactant:
169 g . 1mol /481.77g = 0.351 moles of RuI₃
58g . 1mol /28g = 2.07 moles of CO
96.2g . 1mol/ 107.87g = 0.892 moles
Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.
3 moles of Ag react to 1 mol of RuI₃
Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles
We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.
1 mol of RuI₃ react to 3 moles of Ag
Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles
It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.
5 moles of CO react to 3 moles of Ag
Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.
This calculate confirms the theory.
Now, we determine the maximum mass of Ru(CO)₅
3 moles of of Ag can produce 1 mol of Ru(CO)₅
Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles
We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g
A student uses 200 grams of water at a temperature of 60 °C to prepare a saturated solution of potassium chloride , KCI. Identify the solute in this solution.
1. H2O(l)
2. KCl (aq)
3.K + (aq)
4.KCl(s)
Answer:
4. KCl(s)
Explanation:
KCl is an ionic salt that dissolves in water to form a KCl aqueous solution.
A solution is defined as the homogeneous mixture of one or more solutes dissolved in a solvent. Here in the saturated solution of potassium chloride, the solute is KCl. The correct option is 4.
What is a solute?A solute is defined as the substance which is dissolved in a solution. In a solution the amount of the solute present is always smaller than the amount of the solvent. For example in a salt solution, salt dissolves in water and therefore salt is the solute.
The particles of the solute present in a solution cannot be seen by our eye. The solute from a solution is not possible to separate by filtration. In an unsaturated solution, the concentration of the solute is much lower than that of the concentration of the solvent.
A solution is a combination of the solute and the solvent.
Thus the correct option is 4 - KCl.
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find the concentration in mol/dm^3 of a solution of sodium hydroxide if it contains 3.5g of NaOH in 100cm^3 of solution.
Answer:
0.875
...............
The reaction for the formation of ammonia is given a
N2(g) + 3H2(g) = 21H3(g)
write the vate reaction for :
a) the formation of NH₃.
6) the disappearance of N and H₂
Answer:
a.
[tex]rate_{NH_3}=\frac{2d[NH_3]}{dt}[/tex]
b.
[tex]rate_{N_2}=\frac{-1d[N_2]}{dt} \\\\rate_{H_2}=\frac{-3d[H_2]}{dt}[/tex]
Explanation:
Hello there!
In this case, according to the law of rate proportions, it is possible to write the the rates of reaction for the formation of NH3 and the disappearance of the N2 and H2, by considering that the coefficient in the reaction for NH3 is +2 and those of N2 and H2, -1 and -3 respectively. Moreover, we set up these equations as derivatives as shown below:
a.
[tex]rate_{NH_3}=\frac{2d[NH_3]}{dt}[/tex]
b.
[tex]rate_{N_2}=\frac{-1d[N_2]}{dt} \\\\rate_{H_2}=\frac{-3d[H_2]}{dt}[/tex]
Best regards!
What is another name for heat?
A
light energy
B.
infrared energy
C
Thermal energy
D
electromagnetic energy
Answer:
C
Explanation:
thermal energy is the another name for heat
Which has more mass, 1 mole of magnesium or 1 mole of calcium?
Answer:
calcium has more mass
Explanation:
1 mole calcium mass= 40.078 g
1 mole magnesium mass= 24.305 g
A chemical change creates a new
A-atom
B-element
C-substance
Answer:
c substance
Explanation:
Matter is never destroyed or created in chemical reactions. The particles of one substance are rearranged to form a new substance. The same number of particles that exist before the reaction exist after the reaction.
1.Why do you think the agricultural revolution led to more population growth?
Answer:
The agricultural revolution increased agricultural production and technological advancements. I think this led to population growth because the increase in labor and the increase in technologies increased human deveolpment. The revolution also allowed farmers to grow and produce more food and transport it to where it was needed.
Explanation: