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Sophie [7]
3 years ago
6

How many sigfigs does 50 have? pls explain why

Chemistry
1 answer:
Yuri [45]3 years ago
3 0

Answer: 3 sigfigs

Explanation:youre welcome

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An excess of mg(s) is added to 100.ml of 0.400 m hcl. at 0c and 1 atm pressure, what volume of h 2 (g) can be obtained?
ra1l [238]
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1

number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant 
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂

we can use ideal gas law equation to find the volume of H₂
PV = nRT 
where 
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation 

101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL 
therefore answer is 
c. 448 mL 
5 0
3 years ago
How many atoms are in 5.5 moles of carbon dioxide
Veronika [31]

Answer: I think the answer is, 3.312177825e+24 atoms

Explanation: I had a problem similar to this, Hope this helps!

4 0
3 years ago
Read 2 more answers
A block of material has a volume of 50 cubic centimeters, and a mass of 1,000 grams; what is that object's density?
zhannawk [14.2K]

Answer:

Density = 20\ g/cm^3

Explanation:

Given that,

Mass of the object, m = 1000 g

Volume of the block, V = 50 cm³

We need to find the density of the object. Density is equal to mass per unit volume.

d = m/V

d=\dfrac{1000\ g}{50\ cm^3}\\\\=20\ g/cm^3

So, the density of the object is 20\ g/cm^3.

8 0
3 years ago
Use coulomb's law to calculate the ionization energy in kj/mol of an atom composed of a proton and an electron separated by 185.
Tems11 [23]
Coulomb's law mathematically is:
F = kQ₁Q₂/r²
we integrate this with respect to distance to obtain the expression for energy:
E = kQ₁Q₂/r; where k is the Coulomb's constant = 9 x 10⁹; Q are the charges, r is the seperation
Charge on proton = charge on electron = 1.6 x 10⁻¹⁹ C
E = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹) / (185 x 10⁻¹²)
E = 1.24 x 10⁻¹⁸ Joules per proton/electron pair
Number of pairs in one mole = 6.02 x 10²³
Energy = 6.02 x 10²³ x 1.24 x 10⁻¹⁸
= 746.5 kJ
5 0
3 years ago
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Whivh is the correct mole ratio of K3N to KCI in the chemical reaction ​
Firlakuza [10]

3.65. The mole ratio from the balanced equation is 2 moles CO2 : 2 moles CO. 2.

4 0
3 years ago
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