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snow_tiger [21]
3 years ago
5

Use the periodic table in the tools bar to answer these questions. How many moles of AgNO3 are present in 1.50 L of a 0.050 M so

lution?
Chemistry
2 answers:
IgorC [24]3 years ago
4 0
M= moles de soluto / litros de solucion 
moles de soluto = M. litros de solucion 

Moles de soluto = 0.050 M x 1.50 L = 0.075 moles de AgNo3
mixer [17]3 years ago
3 0

m = Solute moles/liters solution moles of solute =

m. Liters of solution moles of solute = 0050 m x 1.50 L = 0075 moles of AgNo3


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Usimov [2.4K]
The answer is 232 plus 450
7 0
3 years ago
A cube of an unknown metal measures 0.200 cm on one side. The mass of the cube is 52 mg. Which of the following is most likely t
Margarita [4]

Answer:

Option E. Zirconium

Explanation:

From the question given above, the following data were obtained:

Length of side (L) of cube = 0.2 cm

Mass (m) of cube = 52 mg

Name of the unknown metal =?

Next, we shall determine the volume of the cube. This can be obtained as follow:

Length of side (L) of cube = 0.2 cm

Volume (V) of the cube =?

V = L³

V = 0.2³

V = 0.008 cm³

Next, we shall convert 52 mg to g. This can be obtained as follow:

1000 mg = 1 g

Therefore,

52 mg = 52 mg × 1 g / 1000 mg

52 mg = 0.052 g

Thus, 52 mg is equivalent to 0.052 g.

Next, we shall determine the density of the unknown metal. This can be obtained as follow:

Mass = 0.052 g.

Volume = 0.008 cm³

Density =?

Density = mass / volume

Density = 0.052 / 0.008

Density of the unknown metal = 6.5 g/cm³

Comparing the density of the unknown metal i.e 6.5 g/cm³ with those given in table in the above, we can conclude that the unknown metal is zirconium

7 0
3 years ago
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Anni [7]

Answer:

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5 0
3 years ago
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Tatiana [17]
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4 0
3 years ago
What volume of 0.500 m h2so4 is needed to react completely with 20.0 ml of 0.400 m lioh?
ExtremeBDS [4]
The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol 
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol 
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol  in 1 L solution 
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L 
therefore volume of acid required = 8 mL 
4 0
3 years ago
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