Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How many moles of hydronium ions H3O are neutralized by 1 teaspoon of antacid product?

Answer:

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Explanation:

Answer:

Water filters remove elements that cause drinking water to have an unpleasant taste and smell, such as lead, chlorine and bacteria. Home water filtration system will improve the overall purity, taste and smell of your drinking water. It also lowers the pH level of the water that you drink.

Answer:

6.684g

Explanation:

Here, we can use the mole ratio of the gases to calculate.

We know that the mole ratio of the gases equate to their number of moles.

Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol

Thus, the number of moles produced is 5.98/32 = 0.186875

Where do we move from here?

We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.

449/851 = 0.186875/n

n =(0.186875 * 851)/449

n = 0.3542

Now we do the same for argon to get the number of moles of argon.

Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.

P = 851 - 449 = 402 mmHg

We now use the mole ratio relation.

402/851 = n/0.3542

n = (402 * 0.3542) / 851

n = 0.1673

Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.

The atomic mass of argon is 39.948 amu

The mass is thus 39.948 * 0.1673 = 6.684g

Answer:

The various uses of water :

1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.

2. Water is used as a universal solvent.

3. water maintains the temperature of our body.

4. Water helps in digestion in our body.

5 .water is used in factories and industries.

6. Water is used to grow plants , vegetables and crops.

Answer:

The various use of water are;

I) Cooking.

ii) Drinking

III) Bathing

iv) Generating hydro- electricity

v) Construction work etc

Answer:

LiOH(aq) → Li⁺(aq) + OH⁻(aq). 

Answer:

[tex]C_t=0.165M[/tex]

Explanation:

From the question we are told that:

Slope [tex]K=0.056 M-1 s -1[/tex]

initial Concentration [tex]C_1=2.2M[/tex]

Time [tex]t=100[/tex]

Generally the equation for Raw law is mathematically given by

[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]

[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]

[tex]C_t=0.165M[/tex]

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]

Substitute values in the above equation:

[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]

Therefore, after 100 seconds the concentration is 0.165 M.

Learn more about the order of reaction here:

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Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.

At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.

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Explanation:

here's the answer to your question

a].When sand is added to water it either hangs in the water or forms a layer at the bottom of the container. Sand therefore does not dissolve in water and is insoluble. It is easy to separate sand and water by filtering the mixture.

b]. The water filtered by kamal is not safe to drink .

If you run out of water, or cannot carry enough water with you for your entire trip, you may need to source drinking water from natural water sources.

Answer

A

Explanation:

due to high specific heat capacity it loses heat and has low temperature

Answer:

26mL

Explanation:

NaOH+HCl= NaCl+H2O

nHCl=0.0243*2=0.0486

nNaOh=nHCl

VNaOH=0.0486/1.87=0.026l=26ml

Answer:

26.0 mL

Explanation:

Answer:

a. -58 millivolts

Explanation:

The given Nernst equation is:

[tex]E_{ion} = 58 millivolts /z \Big[ log_{10} \Big( \dfrac{[ion]_{out}}{[ion]_{in}}\Big) \Big]}[/tex]

The equilibrium potential given by the Nernst equation can be determined by using the formula:

[tex]E_{Cl^-} = \dfrac{2.303*R*T}{ZF} \times log \dfrac{[Cl^-]_{out}} {[Cl^-]_{in}}[/tex]

where:

gas constant(R) = 8.314 J/K/mol

Temperature (T) = (20+273)K

= 298K

Faraday constant F = 96485 C/mol

Number of electron on Cl = -1

[tex]E_{Cl^-} = \dfrac{2.303*8.314*298} {(-1)*(96845)} \times log \dfrac{100} {10}[/tex]

[tex]E_{Cl^-} = - 0.05814 \ volts[/tex]

[tex]\mathsf{E_{Cl^-} = - 0.05814 \times 1000 \ milli volts}[/tex]

[tex]\mathsf{E_{Cl^-} \simeq - 58\ milli volts}[/tex]

Answer:

However, there has to be 2 electrons on the first shell, and 8 on the others.

Explanation:

Hope this helps :)

Answer:

1.28 M

Explanation:

Step 1: Given data

Mass of luminol (solute): 17.0 g

Volume of water: 75.0 mL (this is also the volume of solution)

Step 2: Calculate the moles corresponding to 17.0 g of luminol

The molar mass of luminol is 177.16 g/mol.

17.0 g × 1 mol/177.16 g = 0.0960 mol

Step 3: Calculate the molarity of the solution

We will use the definition of molarity

M = moles of solute / liters of solution

M = 0.0960 mol / 0.0750 L = 1.28 M

Answer:

35

Explanation:

is the answer for your question

Answer:

19.2 g

Explanation:

Step 1: Write the balanced equation

P₂O₅(s) + 3 H₂O(l) ⇒ 2 H₃PO₄

Step 2: Calculate the moles corresponding to 5.29 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

5.29 g × 1 mol/18.02 g = 0.294 mol

Step 3: Calculate the theoretical yield of phosphoric acid, in moles

The molar ratio of H₂O to H₃PO₄ is 3:2. The theoretical yield of H₃PO₄ is 2/3 × 0.294 mol = 0.196 mol

Step 4: Calculate the mass corresponding to 0.196 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

0.196 mol × 97.99 g/mol = 19.2 g

Answer:

I think its A maybe am not sure

Answer:

790 years

Explanation:

Given that;

0.693/t1/2 = 2.303/t log [A]o/[A]

So;

t1/2 =half life of carbon-14

t= age of the sample

[A]o= activity of the living sampoke

[A] = activity at time t

0.693/5.73 ×10^3 = 2.303/t log 85/77

1.21 × 10^-4 = 2.303/t log 1.1

1.21 × 10^-4 = 0.0953/t

t= 0.0953/1.21 × 10^-4

t= 790 years (to 2sf)

Answer:

D

Explanation:

We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.

Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.

Hence, for this solution, the concentration of ions in solution follows the order;

[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]

Answer:

hello? are you still here? reply if you are

Answer:

D. [CO₂]

Explanation:

Let's consider the following equation at equilibrium.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients. It only includes gases and aqueous species.

Kc = [CO₂]

Answer:

[tex]T_2= 36.7 \textdegree C[/tex]

Explanation:

Mass of Water [tex]m_w=6.90kg[/tex]

Temperature [tex]T=34.7 degrees[/tex]

Heat Flow [tex]H=57.1kJ[/tex]

Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]

Generally the equation for Final Temperature is mathematically given by

[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]

[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]

Therefore

[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]

[tex]T_2= 36.7 \textdegree C[/tex]

Answer:

I say Star S has radial motion

Explanation:

I'm not sure if it right but let me know if you have any other questions

Answer:

a. 4.41 g of Urea

b. 1.5 g of Urea

Explanation:

To start the problem, we define the reaction:

2NH₃ (g) +  CO₂ (g) → CH₄N₂O (s)  +  H₂O(l)

We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:

2.6 g . 1mol / 17g = 0.153 moles of ammonia

Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea

0.153 moles ammonia may produce, the half of moles

0153 /2 = 0.076 moles of urea

To state the theoretical yield we convert moles to mass:

0.076 mol . 58 g/mol = 4.41 g

That's the 100 % yield reaction

If the percent yield, was 34%:

4.41 g . 0.34 = 1.50 g of urea were produced.

Formula is (Yield produced / Theoretical yield) . 100 → Percent yield

Explanation:

the force of the lone pairs from the bottom would cancel out the force of the lone pairs from the top. Thus, the molecule will be linear.

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

The concentration of hydroxide ion of given solution is 4.3 x 10⁻¹¹M.

We can calculate the concentration of hydroxide ions as follow:

[OH⁻][H⁺] = 10⁻¹⁴

Given chemical reaction with ICE table shown as below:

                H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq)

Initial:             0.115                      0                    0

Change:           -x                        +x                  +x

Equilibrium:  0.115-x                   +x                  +x

Given that, Ka = 4.3 x 10⁻⁷

Equilibrium constant for this reaction is written as:

Ka = [H⁺][HCO₃⁻]/[H₂CO₃]

4.3 x 10⁻⁷ = x² / 0.115

x = 2.32 x 10⁻⁴M = [H⁺]

Now we calculate the concentration of hydroxide ion as:

[OH⁻][H⁺] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴ / 2.32 x 10⁻⁴ = 4.3 x 10⁻¹¹M

Hence, value of [OH⁻] is 4.3 x 10⁻¹¹M.

To know more about pH & pOH, visit the below ink:

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Answer:

is this question or you just asking I can't understand.

Answer:

21.5mL of a 0.100M HCl are required

Explanation:

The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.

To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:

Mass NaC6H5O:

1.000g * 25% = 0.250g NaC6H5O

Moles NaC6H5O -116.09g/mol-

0.250g NaC6H5O * (1mol/116.09g) = 2.154x10⁻³ moles = Moles of HCl required

Volume 0.100M HCl:

2.154x10⁻³ moles HCl * (1L/0.100mol) = 0.0215L =

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