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2 years ago

Which of these elements has two valence electrons? A.hydrogen B.barium C.nitrogen D.krypton E.bromine

Chemistry

1 answer:

Umnica [9.8K]2 years ago

3 0

Answer:

Barium

Explanation:

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Explain why a fluorescent light bulb is not as hot as an incandescent light bulb.

Jet001 [13]

An incandescent bulb becomes hotter than a fluorescent bulb when turned on because in a regular incandescent bulb, there is tungsten wire where electricity is converts into heat. A regular incandescent light bulb requires 4 times more energy than a fluorescent bulb in order to produce the same amount of light. The conversion is such that for a 75-watt bulb, temperature get raised to approximately 2000 K. For such a high temperature, the radiating energy from the wire have some visible light. In such bulbs, 90% of the electricity get consumed in producing heat and only 10% produces light thus, they are not much efficient source of light.

On the other hand, fluorescent bulbs produce light with less amount of heat. In them, 40% of electricity is consumed in producing light and 60% in heat which is very less as compared to heat produced by a incandescent bulb. This is because when it get turned on, mercury atoms inside the bulb collides with electrons and produce UV light which is then converted into visible light using thin layer of phosphor power present inside the bulb. This produces low amount of heat thus, the bulb stays cooler, the bigger size of bulb also helps in dispersing heat.

Therefore, a fluorescent light bulb is not as hot as an incandescent light bulb.

5 0

3 years ago

Read 2 more answers

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

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3 years ago

Judging by the electron domain geometry, which type of hybrid orbitals does this molecule have? s sp sp2 sp3

Aleksandr-060686 [28]

The molecule have TETRAHEDRAL HYBRID ORBITAL.
SP3 hybridization involves the mixing of one orbital of S sub level and three orbitals of P sub level of the valence shell. All the orbitals possess equivalent energies and shapes. The SP3 orbital has 25% S character and 75% P character. S and P refers to the s and p sub shells.<span />

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2 years ago

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How are planes contrail lines similar to clouds

Tom [10]

Both are mainly composed of droplets of condensed water

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3 years ago

In the city a street peddler offers you a diamond ring for 30 bucks how can you test the rock

larisa86 [58]

Explanation:

Hardness test — Scratch the rock with a fingernail, a copper penny, a glass plate or nail, and a ceramic plate. Check your Guide to assign it a rating on the Mohs Scale of Hardness.

Color streak test — Test for the “color streak” of the minerals by rubbing the rock across the ceramic plate in the Mineral Test Kit, or across smooth

cement. Look up which colors indicate which minerals are present.

Magnetism test — Hold the magnet in the Mineral Test Kit near your rock. If there is a magnetic pull, it has a metal mineral in it.

Acidity test — Put vinegar in the bottle included in the Mineral Test Kit. Squeeze out a few drops on the rock. If it fizzes, it contains carbonate.

A quick and easy way to find out whether your diamond is real or fake: try fogging it up with your breath. If it clears up after one or two seconds, then your diamond is real, but if it stays fogged for three to four seconds chances are that you're looking at a fake.

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3 years ago