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Stolb23 [73]
4 years ago
11

Can someone help, please ??

Chemistry
1 answer:
mart [117]4 years ago
5 0
9. all you have to do find the atomic mass of each atom in the periodic table.

Na is 23.0
O= 16.0

molecular mass of Na2O= 2x23.0 + 16.0= 62.0 g/mol

10. this is the description of an acid


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s344n2d4d5 [400]
Speed is how fast something is going, velocity is speed and distance, and acceleration is how fast something is speeding up.
4 0
4 years ago
if 3.0 grams of aluminum and 6.0 grams of bromine react to form AlBr3, how many grams of product would theoretically be produced
Gelneren [198K]
1) Chemical reaction: 2Al + 3Br₂ → 2AlBr₃.
m(Al) = 3,0 g.
m(Br₂) = 6,0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 3,0 g ÷ 27 g/mol.
n(Al) = 0,11 mol.
n(Br₂) = n(Br₂) ÷ m(Br₂).
n(Br₂) = 6 g ÷ 160 g/mol.
n(Br₂) = 0,0375 mol; limiting reagens.
n(Br₂) : n(AlBr₃) = 3 : 2.
n(AlBr₃) = 0,025 mol.
m(AlBr₃) = 0,025 mol · 266,7 g/mol.
m(AlBr₃) = 6,67 g.

2) m(Br₂) - all bromine reacts, so mass of bromine after reaction is zero grams (m(Br₂) = 0 g).
n(Al) = 0,11 mol - 0,025 mol = 0,085 mol.
m(Al) = 0,085 mol · 27 g/mol.
m(Al) = 2,295 g.
m(AlBr₃) = 6,67 g · 0,72 (yield of reaction).
m(AlBr₃) = 4,8 g.
n - amount of substance.
M - molar mass.

4 0
3 years ago
If 1.2 moles of ideal gas occupy a volume of 18.2 L at a pressure of 1.8 ATM what is the temperature of gas in degrees Celsius.
Novosadov [1.4K]

Answer: A 59.5 degree celcius

The equation that we will use to solve this problem is :

PV = nRT where:

P is the pressure of gas = 1.8 atm

V is the volume of gas = 18.2 liters

n is the number of moles of gas = 1.2 moles

R is the gas constant =  0.0821

T is the temperature required (calculated in kelvin)

Using these values to substitute in the equation, we find that:

(1.8)(18.2) = (1.2)(0.0821)(T)

T = 332.5 degree kelvin

The last step is to convert the degree kelvin into degree celcius:

T = 332.5 - 273 = 59.5 degree celcius

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Except for the Middle East :)
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Does the shape of molecule matter?​
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Yes because it helps determine the properties the molecule has.
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