Answer:
437.5 mL
Explanation:
Concentration (c) = Moles (n) / Volume of solution (liters)
C = n / V
rearrange the formula to find out how many moles you need in the final solution.
n (moles) = concentration (c) x Volume (V) ( in liters)
Convert your mL into L. 1500mL is 1.5L
n = 3.5 moles/liter x 1.5 liters (the liters cancel each other out)
n = 5.25 moles
Now use the same formula to calculate the volume you need to make the final solution based on the concentration of the original.
c = n/V or V = n / c
V = 5.25 moles / 12 moles/liter here the moles cancel out
V = 0.4375 L
convert back to mL
V = 437.5 mL
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.
what is a heterogeneous mixture?
Answer:
A heterogeneous mixture is a mixture in which the composition is not uniform throughout the mixture
Which element does this Bohr model represent? (look at a periodic table if needed)
Ive finished my class good luck!!? ❤️✌️
Answer:
yayyy! proud of u
Explanation:
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
Plz help me I am timed!!!!!
Answer:
i think its the gibbous phase
Explanation:
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:
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
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]
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What happened during the fusion reaction shown? 2H Зн 4He neutron O A. Two H nuclei fused into one He nucleus. O B. One He nucleus split into two H nuclei. O C. The nuclei that fused lost all their neutrons. D. The nuclei that fused lost some of their protons.
Answer: C.
Explanation:
two H nuclei fused into one He nucleus
Answer: A!
Explanation: just did it
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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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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Anyone know the answer
Answer:
b
Explanation:
i know
The oxidation state/number for X in XO3- is:
Answer:
We can Observe that the Compound does not have the Positive or the negative Charge thus it is Neutral. Hence, the Oxidation Number of the Compound Will be 0. Thus, the Highest Oxidation Number of the X is 6.
The oxidation state/number for X in XO3 is determined as +6.
Oxidation state of X in XO3
The oxidation state/number for X in XO3 is determined from the overal charge of the compound and the charge of oxygen atom.
Overall charge of the compound = 0Oxidation state of oxygen (O) is -2The oxidation state of "X" is calculated as;
X + (3 x -2) = 0
X - 6 = 0
X = +6
Thus, the oxidation state/number for X in XO3 is determined as +6.
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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
...............
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
Plz help me I am timed!!
Answer:
What do you need help with
Explanation:
Explanation:
okay I'll help but whats the question
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
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!
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.
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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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
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
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
Ozone is three oxygen atoms bonded together. It can sometimes be smelled in the air after a thunderstorm. the smell is a result of
Ozone gas can be smelled in the air after a thunderstorm as a result of reactions between oxides of nitrogen and volatile organic compounds.
How ozone gas be smelled?Ozone gas can be produced due to chemical reactions between oxides of nitrogen and volatile organic compounds. This reaction occurs when pollututed gas is emitted from vehicles react and other activities with each other in the presence of sunlight.
So we can canclude that Ozone gas can be smelled in the air after a thunderstorm as a result of reactions between oxides of nitrogen and volatile organic co
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The correct name for HIO2?
Answer:
iodous acid
Explanation:
iodous acid would also be known as HIO2
10 points pls help!!
Answer:
Question 16: I'm not sure but most likely #3
Question 17: #4
Explanation: Hope this Helps
Answer:
A for first one and second one is A
Using the balanced equation N2+O2=2NO, how many grams of NO can be produced when 25.0 grams of N react?
Answer:
[tex]53.55gNO[/tex]
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:
[tex]=25.0gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNO}{1molN_2}*\frac{30.01 gNO}{1molNO}\\\\=53.55gNO[/tex]
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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.
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