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Nata [24]
2 years ago
15

A typical virus is 5x10-6 cm in diameter. if avogadro's number of these virus particles were laid in a row, how many kilometers

long would the line be?
Chemistry
2 answers:
Debora [2.8K]2 years ago
8 0
Answer is: line be long 3,011·10¹³ kilometers.
diametar of virus = 5·10⁻⁶ cm ÷ 100000 = 5·10⁻¹¹ km.
line lenght = 5·10⁻¹¹ km · 6,023·10²³.
line lenght = 3,011·10¹³ km.
Avogadro number = 6,023·10²³.
1 cm = 10⁻² m = 10⁻⁵ km.
ad-work [718]2 years ago
6 0

Answer:

3.01x10^{12} Km

Explanation:

The avogrado's number is A = 6.02x10^{23}, which means that in one mol of a substance, there is this number of molecules, or atoms, for example.

So, to know how long would be the line (L), we must multiply the numbers of diameter and avogrado:

L = 5x10^{-6} x 6.02x10^{23}

To do it, we must multiply the numbers, repeat the basis 10 and add the exponents, then:

L = 3.01x10^{18} cm 1 cm ------ ------------------10^{-6} Km3.01 x 10^{18} cm --------xBy simple direct rule: x = 3.01x10^{12} Km

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The pressure of a 609.64 gram sample of F2 in a 88.84 L container is measured to be 2770.96 torr. What is the temperature of thi
Oxana [17]

Answer : The temperature of the gas is, 245.9 K

Explanation :

To calculate the temperature of gas we are using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT

where,

P = pressure of gas = 2770.96 torr = 3.646 atm

Conversion used : (1 atm = 760 torr)

V = volume of gas = 88.84 L

T = temperature of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

w = mass of gas = 609.64 g

M = molar mass of F_2 gas = 38 g/mole

Now put all the given values in the ideal gas equation, we get:

(3.646atm)\times (88.84L)=\frac{609.64g}{38g/mole}\times (0.0821L.atm/mole.K)\times (T)

T=245.9K

Therefore, the temperature of the gas is, 245.9 K

4 0
2 years ago
Salim takes 100 g of water each in four identical containers P, Q, R and S and mixes the following in them. 5 g of salt in conta
Simora [160]

The container with chalk powder will contain the least amount of water, because it absorbs water, but the containers with honey and cocunut oil will conserve their amount of water, because they will prevent the water evaporation (especially cocunut oil because it will be on the top side of the container).

7 0
2 years ago
What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
ololo11 [35]
Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995

So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

Have a nice day!
3 0
2 years ago
If it takes three "breaths" to blow up a balloon to 1.2 l, and each breath supplies the balloon with 0.060 moles of exhaled air,
vesna_86 [32]
<span>It takes 3 breaths to get to 1.2 l. One breath is then (1.2 l) / 3 breaths = .4l/breath. To get to 3.0 l we need the difference from 1.2 l. 3.0-1.2 = 1.8 l. Divide the difference by liters/breath (.4) to get how many needed breaths. (1.8 l)/(.4 l/breath) = 4.5 breaths to get the balloon to 3.0 l. In total there were 3 breaths+ 4.5 breaths = 7.5breaths to get to 3.0 l. To find the total moles multiply 7.5breaths by .060 moles/breath 7.5 breaths*.060moles/breath = .45moles</span>
6 0
2 years ago
Read 2 more answers
Cu + 2AgNO3 es002-1.jpg 2Ag + Cu(NO3)2 How many moles of copper must react to form 0.854 mol Ag?
marin [14]
Balance Chemical Equation is as follow,

<span>                        Cu + 2 AgNO</span>₃     →    2 Ag + Cu(NO₃)₂

According to Balance Equation,

                   2 Moles of Ag is produced by reacting  =  1 Mole of Cu
So, 
     0.854 Moles of Ag will be produced by reacting  =  X Moles of Cu

Solving for X,
                             X  =  (0.854 mol × 1 mol) ÷ 2 mol

                             X  =  0.427 Moles of Cu
Result:
            0.854 Moles of Ag 
are produced by reacting 0.427 Moles of Cu.
4 0
2 years ago
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