How do you fight off ADHD medication

Answers

Answer 1
Answer:

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.


Related Questions

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What is impossible for a machine to do?A. do a greater amount of work than the amount of work done on the machineB. apply a force in a direction that is different than the direction of the force applied to the machineC. move an object a greater distance than the distance that part of the machine was movedD. apply a force that is less than the force that is applied to the machine
Which of the following describes a way in which chemistry affects how youuse a computer?A. The colors in the computer display are the result of substancesthat glow.B. By clicking a mouse, you can select an object or text on acomputer screen.C. Eye stress can be reduced by sitting a certain distance from theScreen.D. You are able to play video games by moving your body instead of ajoystick
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A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Four different methods are described for validating the results of a particular analysis. Indicate for each whether the method primarily checks the accuracy of the analysis or the precision of the analysis.a. Five aliquots of the same sample are injected for gas chromatographic analysis by one person on the same day.
i. Accuracy
ii. Precision
b. A known amount of analyte is added to an aliquot of the sample and analyzed with sample.
i. Accuracy
ii. Precision
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i. Accuracy
ii. Precision
d. Identical standard are analyzed by two different methods.
i. Accuracy
ii. Precision

Answers

Answer:

a) Precision

b) Accuracy

c) Accuracy and precision

d) Accuracy

Explanation:

When an experiment is done more than once to determine if the results are statistically ok, two forms of the validations are possible the accuracy and precision. When the values of the various experiments are close to the known value, then they are accurate. When the values are close to each other they are precise. So, sometimes the results are precise but are not accurate, and vice-versa.

a) Here, the person wants to find if the 5 aliquots will have close results, so, he or she is looking for precision.

b) Here the amount of analyte is already known, and the person wants to identify if the value will be the same when analyzed together with another sample, thus he or she is looking for accuracy.

c) Here the three results will be compared with each other (precision) and with the standard value of the method (accuracy).

d) The methods will be tested, and the values will be compared with the standard known value, so the person is looking for accuracy.

How many water molecules are in a block of ice containing 1.50 mol of water (H2O)?

Answers

1.50 moles H20(6.02*10^23 molecules/1 mole)
9.03*10^23 molecules

Final answer:

The number of water molecules in a 1.50 mol block of ice is calculated by multiplying the number of moles of water by Avogadro's number. The result is approximately 9.033 x 10^23 water molecules.

Explanation:

In chemistry, the amount of substance in moles is related to the number of particles (atoms, molecules) through Avogadro's number. Avogadro's number, which is 6.022 x 1023 particles/mol, tells us the number of molecules in one mole of a substance.

To calculate the number of water molecules in 1.50 mol of water, you would multiply the number of moles of water by Avogadro's number:

1.50 mol of water x 6.022 x 1023 water molecules/mol of water = 9.033 x 1023 water molecules

Therefore, there are approximately 9.033 x 1023 water molecules in a 1.50 mol block of ice.

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Does the density of a liquid depend on its volume? Write your answer as a CER.

Answers

Answer: Yes

Explanation:

Density of a liquid depend on its volume. This is because Density is mass of liquid divided by volume.

Density is inversely proportional to volume.

As density increases, volume decreases and vice versa. The density for water is 1g/ milliliter but it changes with changes in temperature or there are impurities dissolved in it. Ice is less dense that liquid water and it's the major reason it's float because it's volume is inversely proportional to it's density.

How many moles of sulfuric acid are there in 1.50L of a 6.0 M solution

Answers

Answer:

n= 9moles

Explanation:

n=?, C=6M, V= 1.5L

Applying

n= CV

n= 6×1.5= 9moles

In the context of a balanced chemical equation, a____is a number to the lower right of an element or ion within formula and it applies to____. A____is a number to the left of a formula and it applies to_____. Only_____should be added or changed when balancing chemical equations.

Answers

Answer:

Explanation:

The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH4) and two diatomic oxygen molecules (O2) to produce one carbon dioxide molecule (CO2) and two water molecules (H2O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.

Final answer:

In a balanced chemical equation, a subscript is a number to the right of an element indicating the number of atoms in a molecule. A coefficient is a number to the left of a formula indicating the number of molecules. Only coefficients should be altered when balancing equations.

Explanation:

In the context of a balanced chemical equation, a subscript is a number to the lower right of an element or ion within a formula and it applies to the number of atoms of that element in a molecule. A coefficient is a number placed to the left of a formula and it applies to the number of molecules of the entire substance. Only coefficients should be changed when balancing chemical equations because altering subscripts changes the substance itself.

The balanced chemical equation is a symbolic representation of a chemical reaction, where the number of each type of atom is equalized for both the products and reactants, in accordance with the law of conservation of matter.

For example, in the equation 2H₂O, the subscript '2' to the right of 'H' shows that there are two hydrogen atoms in one water molecule, and the coefficient '2' to the left of 'H₂O' means there are two molecules of water, totaling four hydrogen atoms.

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The half-life of a particular element is 280 days. what is its decay constant?

Answers

The formula for decay is
                                      N' = Ne^(-kt)
with N as the initial amount, k as the decay constant, and N' as the final amount. For decay, 280 days is the time it takes for the initial amount N to decay to 1/2N (half its original mass).
                                 1/2A = Ae^(-280k)
                                 1/2    =   e^(-280k)
Solving for k, we will yield
                                 ln(1/2) = -280k
                                          k = ln(2) / 280 day^-1
                                          k ≈ 2.5 × 10^-3 day^-1