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3 years ago

The illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the source?

Chemistry

2 answers:

Elza [17]3 years ago

7 0

The correct answer is:

D. 96 candelas

Ghella [55]3 years ago

3 0

Answer: D. 96 candelas

Explanation: Illuminance is defined as the quantity of light incident on a surface per unit of area. It depends on the light’s intensity and its distance from the surface.

E= illuminance on the surface (in lux)= 6 lux

I = the intensity of the light from the source (in candelas) = ?

S= distance from the light source to the surface (in meters) = 4 m

$E=\frac{I}{S^2}$

$I=E\times S^2$

$I=6\times (4)^2=96candela$

Read more about density

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8 0

2 years ago

How many grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl?

neonofarm [45]

1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:

n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol

2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation

[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg

3) Transform mol to mass using the atomic mass:

10.4 mol Mg * 24.3 g/mol = 252.7 g of Mg.

4 0

3 years ago

4. Find the pH at each of the following points in the titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The Ka value is 6.6x10-4 a

yawa3891 [41]

Explanation:

Since HF is a weak acid, the use of an ICE table is required to find the pH. The question gives us the concentration of the HF.

HF+H2O⇌H3O++F−HF+H2O⇌H3O++F−

Initial0.3 M-0 M0 MChange- X-+ X+XEquilibrium0.3 - X-X MX M

Writing the information from the ICE Table in Equation form yields

6.6×10−4=x20.3−x6.6×10−4=x20.3−x

Manipulating the equation to get everything on one side yields

0=x2+6.6×10−4x−1.98×10−40=x2+6.6×10−4x−1.98×10−4

Now this information is plugged into the quadratic formula to give

x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2x=−6.6×10−4±(6.6×10−4)2−4(1)(−1.98×10−4)2

The quadratic formula yields that x=0.013745 and x=-0.014405

However we can rule out x=-0.014405 because there cannot be negative concentrations. Therefore to get the pH we plug the concentration of H3O+ into the equation pH=-log(0.013745) and get pH=1.86

6 0

3 years ago

The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions

musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,

x as 0.115 and y as 0.165

Now the mass of AgCl will be,

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,

278.1 × 0.165 = 45.88 grams.

3 0

3 years ago

At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=4.8×

algol [13]

[OH⁻]= 1 x 10⁻¹⁴ : 4.8 x 10⁻⁴ = 2.083 x 10⁻¹¹

3 0

2 years ago