A photon has a wavelength of 750 no. What is the energy in joules

What is the concentration of hydronium ion in a 0.121 M HCl solution?

kogti [31]

Answer:

C) 0.121 M

Explanation:

HCl + H₂O = H₃O⁺ + OH⁻

.121M .121M

HCl is a strong acid . It will dissociate almost 100 % so the concentration of acid and hydronium ion formed will be equal . It is to be noted that hydronium ion is formed due to association of H⁺ and H₂O . H⁺ is formed due to ionisation of HCl .

So concentrtion of hydronium ion ( H₃O⁺ ) will be .121 M.

7 0

3 years ago

2C4 H10 + 13O2 -→ 8CO2 + 10H2O If 356 moles of O2 are available to react, how many moles of CO2 will fom?

nlexa [21]

Answer:

Answer:

The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which simplifies to 1 : 4.

Explanation:

The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.

The chemical equation given is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.

In this case, they are: 2 mol C₄H₁₀ : 13 mol O₂ : 8 mol CO₂ : 10 mol H₂O

Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:

2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to

1 mol C₄H₁₀ : 4 mol CO₂, or

1 : 2.

Explanation:

5 0

3 years ago

Which element in Group 18 of the Periodic Table has the highest lonization energy?

Kruka [31]

all the elements in group 18 are Nobel gases or inert gases . all the elements such as neon , helium, argon etc. ,their outermost shell is completely filled . The noble gases have the largest ionization energies, reflecting their chemical inertness

4 0

3 years ago

Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the

expeople1 [14]

The question is incomplete, the complete question is:

The solubility of slaked lime, $Ca(OH)_2$ , in water is 0.185 g/100 ml. You will need to calculate the volume of $2.50\times 10^{-3}$ M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of $Ca(OH)_2$ = 0.185 g/100 mL

Volume of $Ca(OH)_2$ = 14.5 mL

Using unitary method:

In 100 mL, the mass of $Ca(OH)_2$ present is 0.185 g

So, in 14.5mL. the mass of $Ca(OH)_2$ present will be = $\frac{0.185}{100}\times 14.5=0.0268g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Ca(OH)_2$ = 0.0268 g

Molar mass of $Ca(OH)_2$ = 74 g/mol

Plugging values in equation 1:

$\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol$

Moles of $OH^-$ present = $(2\times 0.000362)=0.000724mol$

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

By the stoichiometry of the reaction:

Moles of $OH^-$ = Moles of $H^+$ = 0.000724 mol

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = $2.50\times 10^{-3}$

Putting values in equation 2, we get:

$2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL$

Hence, the volume of HCl required is 290mL, the mass of $Ca(OH)_2$ is 0.0268g, the moles of

5 0

3 years ago

A solution has a volume of 2.0 L and contains 36.0 grams of glucose (C6H12O6). If the molar mass of glucose is 180 g/mol, what i

zysi [14]

36.0 g of glucose divided by 180 g/mol = 0.200 moles of glucose

find molarity
0.200 moles of glucose / 2 liters = 0.100 molar solution

(hope this helps)

6 0

3 years ago