Answer:
[tex]\boxed {\boxed {\sf 0.0769 \ M}}[/tex]
Explanation:
We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:
[tex]M_AV_A= M_B V_B[/tex]
In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.
[tex]M_A * 25.0 \ mL = M_BV_B[/tex]
The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.
[tex]M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL[/tex]
We are solving for the molarity of the acid and we must isolate the variable [tex]M_A[/tex]. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.
[tex]\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]
[tex]M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]
The units of milliliters cancel.
[tex]M_A= \frac{0.108 \ M * 17.80 }{25.0 }[/tex]
[tex]M_A= \frac{1.9224}{25.0 } \ M[/tex]
[tex]M_A= 0.076896 \ M[/tex]
The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.
[tex]M_A \approx 0.0769 \ M[/tex]
The molarity of the hydrochloric acid is 0.0769 Molar.
Ortho and para hydrogen are....... a). molecular form. b). Nuclear form. c) allotropic form. d). All
Ortho and para hydrogen are nuclei forms
Two methods by which we can conserve water and water the plants.
Answer:
Two methods which help us to conserve water are:
Sprinkler irrigation system: this irrigation has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.
Drip irrigation system: this irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.
how many moles of H2 and N2 can be formed by the decomposition of 0.145 mol of ammonia, NH3 ?
Answer:
Explanation:
The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.
0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]
The reaction for the decomposition of ammonia is as follows:-
[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]
Calculate the mole of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-
[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]
[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]
Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex] are 0.2175 mol, and 0.0725 mol.
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A buffer is a solution that: Select one: a. Results from mixing a strong acid and a strong base. b. When added to another solution, decreases the pH. c. When added to another solution, increases the pH. d. Prevents a drastic change in pH when an acid or base enters a solution.
Answer:
d. Prevents a drastic change in pH when an acid or base enters a solution.
Explanation:
The purpose of a buffer is to resist pH change and keep the solution relatively stable.
Assume that all products containing “Bromide” or an ingredient ending in “-ol” are toxic.
Product B and E are toxic because they contain "BROMIDE" or ingredients that end in 'ol'
Based on the directions given in the information of this question, any product containing "bromide" or containing an ingredient that ends in "ol" is assumed to be TOXIC.
After carefully evaluating the ingredient contents of each product in the image attached to this question, it was realized that product B contains "pyridostigmine bromide" as an ingredient while product E contains "butorphanol" as an ingredient. Hence, in accordance to the guide given in this question, products B and E are toxic.
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If you drip an ink drop into a cup of water and wait for a few seconds, all the water will be colored with the ink. This experiment is an example of facilitated diffusion ?
true
false
Answer:
false, it is not an example of facilitated diffusion
Answer:
TrueExplanation:
When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.
Hydrogen chloride decomposes to form hydrogen and chlorine, like this:
2HCl(g) + H2(g) â Cl2(g)
Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, hydrogen, and chlorine has the following composition:
compound pressure at equilibrium
HCl 84.4 atm
H2 77.9 atm
Cl2 54.4
Required:
Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.
Solution :
Given :
Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm
Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm
Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm
Reaction :
[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]
Using equilibrium concept,
[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]
[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]
[tex]$k_p=0.594$[/tex]
[tex]k_p=0.59[/tex] (in 2 significant figures)
or [tex]k_p=5.9 \times 10^{-1}[/tex]
When a marble is dropped into a beamer of water
Answer:
The water will rise.Explanation:
hope this helps you
-Sweety<3The mass of the marble is greater than that of the water. The marble weighs more than an equivalent volume of the water. The force from dropping the marble breaks the surface tension of the water. The marble has greater mass and volume than the water.
Write the chemical formula for the following:
a. The conjugate acid of amide ion, NH₂-
b. The conjugate base of nitric acid, HNO₃
Follow the rules of Bronsted Lowry theory
Acids take a HBases donate a HSo
#a
NH_2-
Add a H
Conjugate acid is NH_3#b
HnO_3
Take a H
Conjugate base is NO_3-#1
Conjugate acid means one H+ is added
NH_2+H+=NH_3sw
#2
Conjugate base means donate a H+
HNO_3-H=NO_3-A chemist is preparing to carry out a reaction that requires 5.75 moles of hydrogen gas. The chemist pumps the hydrogen into a 10.5 L rigid steel container at 20.0 °C. To what pressure, in kPa, must the hydrogen be compressed? (Show all work for full credit and circle your final answer) *
Answer:
The hydrogen must be compressed to 1333.13302 kPa.
Explanation:
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
P= ?V= 10.5 Ln= 5.75 molesR= 0.082 [tex]\frac{atm*L}{mol*K}[/tex]T= 20 C= 293 K (being 0 C= 273 K)Replacing:
P* 10.5 L= 5.75 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 293 K
Solving:
[tex]P=\frac{5.75 moles* 0.082 \frac{atm*L}{mol*K} * 293 K}{10.5 L}[/tex]
P= 13.157 atm
If 1 atm is equal to 101.325 kPa, then 13.157 atm is equal to 1333.13302 kPa.
The hydrogen must be compressed to 1333.13302 kPa.
A cyclopropane-oxygen mixture is used as an anesthetic. If the partial pressure of cyclopropane in the mixture is 330 mmHg and the partial pressure of the oxygen is 1.0 atm, what is the total pressure of the mixture in torr
Answer:
1090 Torr
Explanation:
Step 1: Given data
Partial pressure of cyclopropane (pC₃H₆): 330 mmHg (330 Torr)Partial pressure of oxygen (pO₂): 1.0 atmStep 2: Convert pO₂ to Torr
we will use the conversion factor 1 atm = 760 Torr.
1.0 atm × 760 Torr/1 atm = 760 Torr
Step 3: Calculate the total pressure of the mixture (P)
The total pressure of the mixture is the sum of the partial pressures of the gases.
P = 330 Torr + 760 Torr = 1090 Torr
J00
Sugar
(C2H2011)
260
KNO
220
180
Solubility (g solute per 100 g H,0)
140
NaNO,
NaBr
100
KBr
60
КСІ
Naci
20
0
0
20
Ce (50)
40 60
Temperature (°C)
80
100
Which compound would make a saturated solution if 98 grams were
dissolved in 100 grams of solution at 80 degrees Celsius?
O KBr
O Sugar
OKCI
O NaCl
alish
how is the molecule of substance formed
Answer:
When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.
Explanation:
Three important nutritional additions to training for a long distance race are
O Protein
O Water
O Increase calories
O All of the answer choices
Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in % wt/v concentration.
Known; Mass of CaCl2 present in original solution, based on actual yield= 1.77g moles
CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles
Total Volume of solution =V, which is 80ml
Answer:
2.21% wt/v
Explanation:
The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL
The %wt/v of this solution is:
%wt /v = 1.77g / 80mL * 100
%wt/v = 2.21% wt/v
The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
Answer:
Moles of water are 0.868
Explanation:
a 82.6 L sample of gas exerts 350.8 mm hg pressure at 134.6 c°. what valine does the gas have at 736.4 mm hg and 42.8 c°
a. 30.5
b 4.13
c. 12.5
d. 134
need help ASAP
Answer:
nfururhrj waltz quiz amora7ersgdsYsdi6 whiz 53
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Consider the following titration for these three questions:
1.00 L of 2.00 M HCl is titrated with 2.00 M NaOH.
a. How many moles of acid are equal to one equivalent in this titration?
b. How many moles of HCl are found in solution at the halfway point of the titration?
c. How many liters of base will be needed to reach the equivalence point of the titration?
Answer:
a. 1 mole of acid is equal to one equivalent.
b. 1.00 moles of HCl are found.
c. 1L of 2.00M NaOH is needed to reach the equivalence point
Explanation:
HCl reacts with NaOH as follows:
HCl + NaOH → NaCl + H2O
Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1
a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent
b. The initial moles of HCl are:
1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl
At the halfway point, the moles of HCl are the half, that is:
1.00 moles of HCl are found
c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:
2.00moles NaOH * (1L / 2.00mol) =
1L of 2.00M NaOH is needed to reach the equivalence point
write anode and cathode in Zn-Ag galvanic cell
Explanation:
Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).
By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).
Calculate the volume of a 89.51 g sample of carbon dioxide at 281.8 K and 843.9 torr. Round your answer
to the nearest L. Do not include units.
Answer:
1,000.000
Explanation:
Leo carefully pipets 50.0 mL of 0.500 M NaOH into a test tube. She places the test tube
into a small beaker to keep it from spilling and then pipets 75.0 mL of 0.250 M HCl into
another test tube. When Leo reaches to put this test tube of acid into the beaker along
with test tube of base she accidentally knocks the test tubes together hard enough to
break them and their respective contents combine in the bottom of the beaker. Is the
solution formed from the contents of the two test tubes acidic or basic? What is the pH of
the resulting solution?
Please answer below questions one by one to assist you receive full credits
(Alternatively, you can discard my hints below, solve the problem using your own way
and send me the picture/copy of your complete work through email)
The mole of NaOH before mixing is
mol (save 3 significant figures)
The mole of HCl before mixing is
mol (save 4 significant figures)
After mixing, the solution is
(choose from acidic or basic)
The total volume of mixture is
L (save 3 significant figures)
The concentration of [OH-] is
M (save 3 significant figures)
The concentration of [H'l is
M (save 3 significant figures)
Let's consider the neutralization reaction between HCl and NaOH.
NaOH + HCl ⇒ NaCl + H₂O
To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.
NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol
HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol
Now, let's determine the reactant in excess and the remaining moles of that reactant.
NaOH + HCl ⇒ NaCl + H₂O
Initial 0.0200 0.0188
Reaction -0.0188 -0.0188
Final 1.20 × 10⁻³ 0
The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³ moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:
[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M
NaOH is a strong base according to the following equation.
NaOH ⇒ Na⁺ + OH⁻
The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.
The pOH is:
pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02
We will calculate the pH using the following expression.
pH = 14.00 - pOH = 14.00 - 2.02 = 11.98
The pH is 11.98. Since pH > 7, the solution is basic.
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what volume will it occupied 40 degrees celsius a gas sample was collected when a temperature is 27 degrees celsius and the volume of 1 L
Using Charles law
[tex]\boxed{\sf \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}[/tex]
[tex]\\ \sf\longmapsto V_1T_2=V_2T_1[/tex]
[tex]\\ \sf\longmapsto 1(40)=27V_2[/tex]
[tex]\\ \sf\longmapsto V_2=\dfrac{40}{27}[/tex]
[tex]\\ \sf\longmapsto V_2=1.48\ell[/tex]
According to the EPA Lead and Copper Rule (LCR), the action level for Pb in drinking water (the level at which threat to human health requires public notification and action towards mitigation) is 15 ppb. If you were to add enough phosphate to the system
saturated with respect to Pb3(PO4)2(s), would the [Pb2+] be below the action limit?
Answer:
The right answer is "105.17 ppb".
Explanation:
According to the question,
The amount of [tex]Pb^{2+}[/tex] in ppb will be:
= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]
= [tex]105.17 \ ppb[/tex]
Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.
An ionic bond is a bond
Answer:
That involve the complete transfer of an electron from one atom of an element to another
A 18.0 L gas cylinder is filled with 6.20 moles of gas. The tank is stored at 33 ∘C . What is the pressure in the tank?
Express your answer to three significant figures and include the appropriate units.
Answer:
8.65 atm
Explanation:
Using ideal law equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (Latm/molK)
T = temperature (K)
According to the information given in this question;
V = 18.0 L
n = 6.20 moles
R = 0.0821 Latm/molK
T = 33°C = 33 + 273 = 306K
P = ?
Using PV = nRT
P × 18 = 6.20 × 0.0821 × 306
18P = 155.76
P = 155.76/18
P = 8.65 atm
Consider the preparation of methyl benzoate by reacting benzoic acid with methanol using sulfuric acid as a catalyst. Reaction scheme of benzoic acid with methanol, conc. sulfuric acid, and heat over the arrow, and methyl benzoate and water as products. Calculate the molar masses of the reactant and product. Report molar masses to 1 decimal place. Molar mass of benzoic acid g/mol Molar mass of methyl benzoate
Answer:
See explanation
Explanation:
The molecular mass is the sum of the relative atomic masses of all the atoms in the molecule.
The relative atomic mass of reactants and products are calculated as follows;
Benzoic acid is C7H6O2 hence the molar mass of benzoic acid is ;
7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122.0 g/mol
Methyl benzoate is C8H8O2
8(12) + 8(1) + 2(16) = 96 + 8 + 32 = 136.0 g/mol
A fusion reaction releases energy because the binding energy of the resulting nucleus:______.
a. is released in the process.
b. is equal to the binding energy of the original nuclei.
c. is absorbed in the process.
d. is less than the binding energy of the original nuclei.
e. is greater than the binding energy of the original nuclei.
Answer:
a. is released in the process
Explanation:
In fusion reaction the nucleus is unstable so it releases its binding energy resulting in decreasing its mass so it becomes more stable.
Is Water and kerosine a mixture
Answer:
No.Kerosene oil and water do not mix with each other and form two separate layers.
Answer:
No
Explanation:
They cannot be mixed together they will form upper and lower layer
9.7300x10^2 + 9.8700x10^3
? × 10^?
Answer:
19.6 x 10⁵
1.96 x 10⁶
I hope it's helps you
Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
21.4g of HBr is the minimum mass that could be left over.
Explanation:
Based on the reaction:
HBr + NaOH → NaBr + H2O
1 mole of HBr reacts per mole of NaOH
To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:
Moles NaOH -40.0g/mol-
17g * (1mol/40.0g) = 0.425 moles NaOH
Moles HBr -Molar mass: 80.91g/mol-
55.8g * (1mol/80.91g) = 0.690 moles HBr
The difference in moles is:
0.690 moles - 0.425 moles =
0.265 moles of HBr could be left over
The mass is:
0.265 moles * (80.91g/mol) =
21.4g of HBr is the minimum mass that could be left over.