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Liula [17]
3 years ago
15

What kind of energy can be mainly excited by us/visible light

Chemistry
1 answer:
erica [24]3 years ago
3 0
I'm thinking the answer you are looking for is light energy. With the sunlight there is light energy and heat.

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What is the [OH-] of solution with pH of 3.4?
arsen [322]

Answer:

(A) 4×0.0001M

Explanation:

[OH-] = 10^-3.4 which is equivalent to 4×0.0001M

5 0
3 years ago
Read 2 more answers
Of which tissue is the dermis composed of?
Maslowich
The dermis or corium is a layer of skin between the epidermis
6 0
3 years ago
If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

8 0
3 years ago
The most common form if matter in the universe is____?
Rashid [163]

I am almost positive that the answer is: Plasmas. Let me know if im right or not please!! :)

7 0
3 years ago
An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi
arsen [322]

Explanation:

In a magnetic field, the radius of the charged particle is as follows.

             r = \frac{mv}{qB}

where,   m = mass,      v = velocity

              q = charge,    B = magnetic field

Therefore, q will be calculated as follows.

         q = \frac{mv}{rB}

            = \frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}

            = \frac{34.98 \times 10^{-22}}{0.01092}

            = 3.2 \times 10^{-19} \times \frac{1.0 e}{1.6 \times 10^{-19}C}

            = +2e

Thus, we can conclude that the charge of the ionized atom is +2e.

5 0
2 years ago
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