Answer:
D 27.74%
Explanation:
Percent composition of Magnesium (Mg) in Magnesium Phosphate compound ie Mg3(PO4)2 = 27.74%
Answer:
The answer is D
Explanation:
I took the test
Heat will continue to move until the objects or areas have reached the same ______.
Answer: Thermal Equilibrium
Explanation:
Answer:
temperature
Explanation:
I think that's it i'm sorry if i'm wrong
what is a row of elements across the periodic table called
Answer:
a period
Explanation:
Answer: The row of elements across the periodic table is called "periods".
Explanation:
In the Periodic Table, there are seven rows of elements, which is called periods.
A chemistry student needs 50.0ml of tetrahydrofuran for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of tetrahydrofuran is . Calculate the mass of tetrahydrofuran the student should weigh out. Be sure your answer has the correct number of significant digits.
Answer:
44.45 g of tetrahydrofuran.
Explanation:
From the question given above, the following data were obtained:
Volume of tetrahydrofuran = 50 mL
Density of tetrahydrofuran = 0.889 g/mL
Mass of tetrahydrofuran =?
Density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, density is expressed as shown below:
Density = mass / volume
With the above formula, we shall determine the mass of tetrahydrofuran needed. This can be obtained as follow:
Volume of tetrahydrofuran = 50 mL
Density of tetrahydrofuran = 0.889 g/mL
Mass of tetrahydrofuran =?
Density = mass / volume
0.889 = mass / 50
Cross multiply
Mass = 0.889 × 50
Mass of tetrahydrofuran = 44.45 g
Therefore, the student should weigh out 44.45 g of tetrahydrofuran.
Given 450.98 g of Cu(NO3)2, how many moles of Ag can be made? Provide your final answer rounded to two decimal places.
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
Answer:
4.82 moles of Ag.
Explanation:
We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:
Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]
= 63.5 + 2[14 + 48]
= 63.5 + 2[62]
= 63.5 + 124
= 187.5 g/mol
Mass of Cu(NO₃)₂ = 450.98 g
Mole of Cu(NO₃)₂ =?
Mole = mass /Molar mass
Mole of Cu(NO₃)₂ = 450.98 / 187.5
Mole of Cu(NO₃)₂ = 2.41 moles
Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:
Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag
From the balanced equation above,
1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.
Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.
Thus, 2.41 moles of Cu is needed for the reaction.
Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:
From the balanced equation above,
1 mole of Cu reacted to produce 2 moles of Ag.
Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.
Thus, 4.82 moles of Ag were obtained from the reaction.