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dimulka [17.4K]
3 years ago
8

How many molecules of CO2 are present

Chemistry
1 answer:
allochka39001 [22]3 years ago
6 0
Where’s the question or equation
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Assume that magnesium would act atom-for-atom the same as copper in this experiment. How many grams of magnesium would have been
Goshia [24]

Answer:

0.1127 grams of Mg is used to produce one gram of silver

Explanation:

Write a balanced equation

Mg(s) +2Ag + (aq) --> Mg^{2+}(aq) + 2Ag(s)

The atomic mass of magnesium is 24.31 g/mol and the atomic mass of silver is 107.84g/mol

1 g Ag of silver is produced from [(1 mol Mg)(24.31 g/mol)] /[ (2mol Ag)(107.84g/mol)]

1 g Ag of silver is produced from 0.1127 grams of Mg

0.1127 grams of Mg is used to produce one gram of silver

7 0
4 years ago
PLEASE HELP ME! What are some patterns or trends that are present in the table of elements?
Sati [7]

Some patterns and trend that are present in the periodic table would be

1. electronegativity (from left-to-right it increases across the table)

2. ionization (from left-to right it increases and from bottom-to-top it increases)

3. electron affinity (same as ionization energy)

4. atom radius (increases opposite way; from right-to-left it increases and from top-to-bottom it increases)

5. melting point (higher melting points with metals and lower melting point with non-metals)

6. metallic character (same as atom radius)

5 0
3 years ago
Radon (Rn) is the heaviest, and only radioactive, member of Group 8A(18) (noble gases). It is a product of the disintegration of
Serggg [28]

Given data                Atomic mass of Ra= 226g/mol

no. of moles =1.0/226g/mol           =0.04424moles

no. of atoms in 0.044moles

no. of atoms =no. of moles x avogadro's number

= 0.044x 6.022 x10^23                  = 0.264968 x 10^22

 If 10^15  atoms of Ra produce 1,373*10^4  atoms of<u> Rn per second</u> then 2,66 *10^21  forms 3,658*10^10 atoms of Rn per second.

Day has 246060=86400 s

That means that 2,66x10^21  atoms of Ra produces 3,16 x10^15  atoms of Rn in a day.

N(Rn)=3.16* 10 ^15                           n(Rn)=N/NA

n(Rn)=5,25*10−9                              pV=nR*T

T=273.15K                                        R=8,314

p=101325Pa                                      V=n∗R∗T/p

V=5.25∗10^−9 ∗ 8.314 ∗ 273.15  /  101325

V=1.1810^−10 m^3 =   118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra

To know more about Ra here

brainly.com/question/9112754

#SPJ4

6 0
2 years ago
Heather just drank 40.0 grams of water (H20). How many moles of water did she just drink? O a. 2.22 moles b. 45 moles O c. 40.0
Ksju [112]

Answer:

This question begins with something, you should know: molar mass from water is aproximately 18 g/m, so if 18 grams of water are contained in 1 mole, the 40 grams occuped 2.22 moles. As you see, opcion a is the best!

Explanation:

5 0
3 years ago
A gas balloon has a volume of 106.0 liters when the temperature is 25.0 °C and the pressure is 740.0 mm Hg. What will its volume
ipn [44]

Answer : The final volume of gas will be, 103.3 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 740.0 mmHg  = 98.4 kPa

Conversion used : (1 mmHg = 0.133 kPa)

P_2 = final pressure of gas = 99.3 kPa

V_1 = initial volume of gas = 106.0 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 25.0^oC=273+25.0=298K

T_2 = final temperature of gas = 20.0^oC=273+20.0=293K

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

\frac{98.4kPa\times 106.0L}{298K}=\frac{99.3kPa\times V_2}{293K}

V_2=103.3L

Therefore, the final volume of gas will be, 103.3 L

8 0
4 years ago
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