As a skier goes down the hill:
A. Gravitational potential energy to kinetic energy
Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
For the steps below, refer to the ICE chart above.
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
Chromic acid, H2CrO4, is a strong oxidizing agent that oxidizes primary and secondary alcohols into carboxylic acids and ketones, respectively. Which technique would be most useful for monitoring the progress of the reaction
Answer:
Thin-layer chromatography
Explanation:
Here in this question, we are concerned with giving the best technique that could be use for monitoring the progress of the reaction that involves the oxidation of alcohols.
The most useful technique here for monitoring the progress of the reaction would be the thin layer chromatography.
Firstly, we should know the reason why we are using a chromatographic technique. This is intuitive. The reason is that the oxidizing agent being used is a colored substance. So basically, the progress of the oxidation reaction would solely rest on the fact that we have a decrease in color saturation of the said oxidizing agent.
This is best done using the thin layer chromatography because we have a mobile phase that moves quickly over the stationary phase and asides this, it moves evenly.
This makes the fact that color changes can be quickly and easily identified and we can tell if the oxidation reaction has gone to completion or not
A pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20
Answer:
C.
Explanation:
Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.
What is the name of Mn(CO3)2
Answer:
Mn is manganese and CO₃ is carbonate. Since the charge for CO₃ is -2 and the subscript is 2, the charge of Mn must be +4 so the answer is manganese (IV) carbonate.
Manganese (IV) carbonate is the name of Mn(CO[tex]_3[/tex])[tex]_2[/tex]. The only names used to identify salts are those of the cation or the anion.
The chemical formula of the anion (such as chloride or acetate) comes first in the name of a salt, which is followed with the identity of the cation (such as sodium or ammonium). They are created when acids and bases react, and they are always composed of either metal cations or cations made of ammonium. Manganese is Mn, and carbonate is CO[tex]_3[/tex]. The solution equals manganese (IV) carbonate since the charge for CO[tex]_3[/tex] is -2 but the subscript is 2, meaning that the charge of Mn has to be +4.
To know more about naming of salt, here:
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Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital?
Answer:
Explanation:
The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital
A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is saying 91g I keep getting 175g. Can someone please explain?
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
Calculate the pH of mixing 24 mL of 1M acetic acid with 76 mL of 1M sodium acetate. For the purpose of this calculation, assume the Ka of acetic acid is 1.8 X 10-5. You must include units to obtain full credit. You must show all your work to obtain any credit.
Answer:
pH = 5.24
Explanation:
Mixture of acetic acid with acetate ion is a buffer (Mixture of a weak acid with its conjugate base). The pH of a buffer can be determined using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka = 4.74; [A⁻] is the concentration of conjugate base (Acetate ion) and [HA] is molar concentration of the weak acid.
Concentration of the acetic acid in the 100mL≡0.1L (76mL + 24mL) solution is:
[HA] = 0.024L ₓ (1mol / L) / 0.1L = 0.24M
[A⁻] = 0.076L ₓ (1mol / L) / 0.1L = 0.76M
Replacing in H-H equation:
pH = 4.74 + log₁₀ [0.76M] / [0.24M]
pH = 5.24
Devise a detailed experimental procedure to purify ~ 20 grams of benzoic acid that is contaminated with sodium chloride. Justification of the steps (including solubility calculations) that are included in the procedure. In other words, explain why the steps are being included.
Answer:
Based on the difference in solubility one can perform the process of purification of the benzoic acid contaminated with sodium chloride. The benzoic acid does not get soluble in cold water, while the sodium chloride is soluble in cold water.
Thus, for separation, the supplementation of cold water can be done into the mixture in the experiment of purifying benzoic acid from sodium chloride. In the process, the mixture is placed on the ice bath and is stirred well, in the end, the solution is filtered. The filtrate contains sodium chloride and on the filter paper pure benzoic acid is collected.
A chemical reaction has the equation 2AgNO3 (aq) + Zn (s) 2Ag (s) + Zn(NO3)2 (aq). What type of reaction occurs between AgNO3 and Zn?
Answer: single replacement reaction
Explanation:
A single replacement reaction is one in which a more reactive element displaces a less reactive element from its salt solution. Thus one element should be different from another element.
A general single displacement reaction can be represented as :
[tex]X+YZ\rightarrow XZ+Y[/tex]
The reaction [tex]2AgNO_3(aq)+Zn(s)\rightarrow 2Ag(s)+Zn(NO_3)_2(aq)[/tex]
When zinc metal is added to aqueous silver nitrate, zinc being more reactive than silver displaces silver atom from its salt solution and lead to formation of zinc nitrate and silver metal.
a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary
Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa), what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K a (C 6H 5COOH) = 6.5 x 10 -5)
Answer:
pH = 4.05
Explanation:
The pH of the benzoic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.187
pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]
Where [] can be understood as moles of each specie.
Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.
Initial moles:
Initial moles of benzoic acid and sodium benzoate are:
Acid: 250mL = 0.250L ₓ (0.250 moles / L) = 0.0625 moles of benzoic acid
Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = 0.050 moles of sodium benzoate
Moles after reaction:
Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:
HCl + C₆H₅COONa → C₆H₅COOH + NaCl
That means after reaction moles of both species are:
Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles
Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles
Replacing in H-H equation:
pH = 4.187 + log [0.0475] / [0.065]
pH = 4.05
Add distilled water to the beaker until the volume
totals 15 mL.
Record the amount of oil that dissolved.
Answer:
i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances
Explanation:
Answer:
None
Explanation:
Answer on Edge 2022
Which of the following best describes a salt bridge? a) A pathway composed of salt water that ions pass through. b) A pathway between the cathode and anode in which ions are reduced. c) A pathway by which counterions can flow between the half-cells with the solutions in the half-cell completely mixing.
Answer: A
Explanation:
Can a catalyst change an exothermic reaction into an endothermic reaction or vice versa? Please explain your answer.
Answer:
A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.
Explanation:
Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.
Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other releases it.
As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.
Select correct answer. The percent yield (isolation yield) of guaifenesin isolated from a 650 mg tablet containing 400 mg dose of drug, can be expressed as: Group of answer choices
Answer:
61.5%.
Explanation:
We usually define percentage yield in chemistry as follows;
Percentage yield= actual yield/theoretical yield ×100
In this case we have;
Actual yield of the guaifenesin 400mg
Theoretical yield of the guaifenesin 650 mg
Hence;
Percentage yield of the guaifenesin =
400/650 ×100 = 61.5%
Hence the theoretical yield of the guaifenesin is 61.5%.
A student uses gravimetric determination to find the water of hydration of a hydrated sample of cobalt (II) sulfate. The student’s sample contained 0.0098459 moles of anhydrous cobalt(II) sulfate (CoSO4) and 0.068921 moles of water (H2O). What is the formula of the cobalt(II) sulfate hydrate? (1 point)
Answer:
CoSO₄.7H₂O
Explanation:
Some salts are in its hydrate form to stabilize them. The hydrated form of CoSO₄ is:
CoSO₄.XH₂O
To find the hydration of the COSO₄ you must find the ratio of H₂O and CoSO₄ that is:
0.068921 moles H₂O / 0.0098459 moles CoSO₄ = 7
That means you have 7 moles of water per mol of CoSO₄ and the formula is:
CoSO₄.7H₂OIf 2 moles of helium undergo a temperature increase of 100 K at constant pressure, how much energy has been transferred to the helium as heat
Answer:
[tex]Q=4154J[/tex]
Explanation:
Hello,
In this case, the involved heat in this heating process is considered to be computed via:
[tex]Q=nCp\Delta T[/tex]
Whereas we assume a constant molar specific heat of helium which is 20.77 J/(mol*K), thus, the transferred energy in the form of heat turns out:
[tex]Q=2mol*20.77\frac{J}{mol*K} *100K\\\\Q=4154J[/tex]
Regards.
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol
The given question is incomplete, the complete question is:
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol
A) –309 kJ/mol
B) –329 kJ/mol
C) None of the above
D) –349 kJ/mol
E) –369 kJ/mol
Answer:
The correct answer is option D, that is, -349 kJ/mol.
Explanation:
Based on the given information, the reaction is:
NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.
Therefore, there is a need to reverse the reaction and change the sign on ΔG.
Now the reaction will become,
Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).
Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 100 mL of water Experiment 2: 4 g of NaOH are dissolved in 200 mL of water Which of the following statements is true?a. both temperature changes will be the sameb. the second temeprature change will be approximately twice the firstc. the second temperature change will be approximately four times the firstd. the second temperature change will be approximately one-half of the firste. the second temperature change will be approximately one-fourth the first
Answer:
a. both temperature changes will be the same
Explanation:
When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:
Q = m×C×ΔT
Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.
Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.
m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.
And Q is the heat released: If 2g release X heat, 4g release 2X.
Thus, ΔT in the experiments is:
Experiment 1:
X / 102C = ΔT
Experiment 2:
2X / 204C = ΔT
X / 102C = ΔT
That means,
a. both temperature changes will be the same
"Calculate the pH during the titration of 20.00 mL of 0.1000 M HF(aq) with 0.2000 M NaOH(aq) after 9.4 mL of the base have been added. Ka of hydrofluoric acid
Answer:
The answer is " 10.39"
Explanation:
Calculating acid moles:
[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]
Calculating NaOH moles:
[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]
calculating excess in OH- Moles:
[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]
calculating total volume:
[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]
[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]
[tex]=0.00025 M[/tex]
[tex]pOH = - \log 0.00025[/tex]
= 3.6
[tex]pH = 14 - pOH[/tex]
= 10.39
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
What is the mole ratio of water to H3PO4?
Answer:
Explanation:
Phosphoric acid H₃PO₄ is produced from reaction of water and tetraphosphorus decoxide P₄O₁₀ as follows
P₄O₁₀ + 6 H₂O = 4 H₃PO₄
In this reaction 6 molecules of H₂O and one molecule of phosphorus compound P₄O₁₀ is needed to produce phosphoric acid , ie the conversion factor of water to acid is 6 / 1
This ratio is called mole ratio of water to H₃PO₄.
So the required ratio is 6 : 1 .
What do we call temperature changes caused by changes in air pressure?
Answer:
Fronts
Explanation:
For example, there are hot and cold fronts which cause the air to become warmer or cooler in a specific region!
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tank contains helium gas at 490 mm Hg, nitrogen gas at 0.75 atm and neon at 520 torr. What is the total pressure in atm? 2.1 atm 0.55 atm 1.5 atm 5.1 atm 51 atm
Answer:
2.1 atm
Explanation:
Before we get the total pressure, we have to ensure all the gases have the same pressure unit.
Nitrogen gas = 0.75 atm
Helium = 490mmHg
To convert mmHg to atm;
760 mmHg = 1 atm
490 = x
x = 460 / 760 = 0.645 atm
Neon = 520 torr
Converting torr to atm;
760 torr = 1 atm
520 torr = x
x = 520 / 760 = 0.684 atm
The total pressure is then given as;
0.75 + 0.684 + 0.645 = 2.1 atm
An aqueous solution of nitric acid is standardized by titration with a 0.110 M solution of calcium hydroxide. If 21.1 mL of base are required to neutralize 23.8 mL of the acid, what is the molarity of the nitric acid solution? ? M Nitric acid
Answer:
[tex]\large \boxed{\text{0.195 mol/L}}[/tex]
Explanation:
(a) Balanced equation
2HNO₃ + Ca(OH)₂ ⟶ Ca(NO₃)₂ + 2H₂O
(b) Moles of Ca(OH)₂
[tex]\text{Moles of Ca(OH)}_{2} = \text{21.1 mL Ca(OH)}_{2} \times \dfrac{\text{0.110 mmol Ca(OH)}_{2}}{\text{1 mL Ca(OH)}_{2}}\\= \text{2.321 mmol Ca(OH)}_{2}[/tex]
(c) Moles of HNO₃
The molar ratio is 2 mol HNO₃:1 mol Ca(OH)₂
[tex]\text{Moles of HNO}_{3} = \text{2.321 mmol Ca(OH)}_{2} \times\dfrac{\text{2 mmol HNO}_{3}}{\text{1 mmol Ca(OH)}_{2}}= \text{4.642 mmol HNO}_{3}[/tex]
(d) Molar concentration of HNO₃
[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V}\\\\c= \dfrac{\text{4.642 mmol}}{\text{23.8 mL}} = \text{0.195 mol$\cdot$L$^{-1}$}\\\\\text{The molar concentration of the Ca(OH)$_{2}$ is $\large \boxed{\textbf{0.195 mol/L}}$}[/tex]
What is the freezing point (in degrees Celcius) of 3.75 kg of water if it contains 189.9 g of C a B r 2?
Answer:
The freezing point of the solution is -1.4°C
Explanation:
Freezing point decreases by the addition of a solute to the original solvent, freezing point depression formula is:
ΔT = kf×m×i
Where Kf is freezing point depression constant of the solvent (1.86°C/m), m is molality of the solution (Moles CaBr₂ -solute- / kg water -solvent) and i is Van't Hoff factor.
Molality of the solution is:
-moles CaBr₂ (Molar mass:
189.9g ₓ (1mol / 199.89g) = 0.95 moles
Molality is:
0.95 moles CaBr₂ / 3.75kg water = 0.253m
Van't hoff factor represents how many moles of solute are produced after the dissolution of 1 mole of solid solute, for CaBr₂:
CaBr₂(s) → Ca²⁺ + 2Br⁻
3 moles of ions are formed from 1 mole of solid solute, Van't Hoff factor is 3.
Replacing:
ΔT = kf×m×i
ΔT = 1.86°C/m×0.253m×3
ΔT = 1.4°C
The freezing point of water decreases in 1.4°C. As freezing point of water is 0°C,
The freezing point of the solution is -1.4°C
Answer:
THE FREEZING POINT IS -1.41 °C
Explanation:
Using the formula of change in freezing point:
ΔTf = i Kf m
i = 3 (1 Ca, 2 Br)
i is the number of the individual elements in the molecules
Kf of water = 1.86 °C/m
mass of CaBr2 = 189.9 g
Calculate the Molar mass of CaBr2:
Molar mass = ( 40 + 80*2) = 200 g/mol
Calculatee the molarity:
molarity = 189.9 g * 1 mole / 200 g/mol / 3.75 kg of water
molarity = 0.2532 M
So therefore, the change in freezing point is:
ΔTf = 1 Kf * M
ΔTf = 3 * 1.86 * 0.2532
ΔTf = 1.41 °C
The freezing point = old freezing point - change in freezing point
The freezing point = 0 - 1.41 °C = - 1.41 °C
The freezing point therefore is -1.41 °C
Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)
Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.
The balanced chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
The heat capacity of liquid water is 4.18 J/g·°C and the heat of vaporization is 40.7 kJ/mol. How many kilojoules of heat must be provided to convert 1.00 g of liquid water at 67°C into 1.00 g of steam at 100°C?
Answer:
The correct answer would be - 2.4KJ or, 2400J
Explanation:
Given:
heat capacity of liquid water - 4.18 J/g·°C
heat of vaporization - 40.7 kJ/mol
Mass of water = 1g
Moles of water = mass/molar mass
= 1g/18.016g
= 0.055 moles
Then,
Total heat required = q1(to raise the temperature to 100) + q2(change from the liquid phase to gas/steam)
= m *s*dt + moles * heat of vaporization
= (1g * 4.18 j/gc * (100-67)) + 0.055* 40.7 KJ
= 137.94J + 2.26KJ
=0.138KJ + 2.26KJ
=2.4KJ or, 2400J
Thus, the correct answer would be - 2.4KJ or, 2400J
The reaction: A + 3 B → D + F was studied and the following mechanism was determined. A + B C (fast) C + B → D + E (slow) E + B → F (very fast) The species, C, is properly described as
Answer:
Intermediate.
Explanation:
Hello,
In this case, we can rewrite the steps as:
[tex]A + B \rightarrow C\ \ (fast)\\\\C + B \rightarrow D + E\ \ (slow)\\\\E + B \rightarrow F \ \ (very fast)[/tex]
Thus, we can notice that in the fast step, C is present as a product but after that is consumed in the slow step, for that reason, and by cause of its formation-consumption behavior, it is properly described as an intermediate as it is not neither a starting-up substance (reactant in the first step) nor a final substance (product in the final step).
Best regards.
how do you fight off ADHD medication
Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.
Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.
It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.