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nignag [31]
3 years ago
6

When the equation Al2(SO4)3 + NaOH + Al(OH)3 + Na2SO4 is correctly balanced, what is the coefficient of Al(OH)3?

Chemistry
1 answer:
nevsk [136]3 years ago
4 0
The balanced chemical formula should be Al2(SO4)3 + 6NaOH = 2Al(OH)3 + 3Na2SO4

Therefore the coefficient of Al(OH)3 is 2!

Hope that helps :)
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A beaker contains 500 ml of 2.40M KNO3 the beaker is left uncovered so that all of the water evaporates. What mass of KNO3 cryst
telo118 [61]

Answer:

242.4 g

Explanation:

RAM of KNO3=39+14+(16×3)=101

Mass=morality×RAM

101*2.4=242.4

6 0
3 years ago
A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi
d1i1m1o1n [39]
The answer is C : 15.7 m/s
Use the idea of : momentum before collision = momentum after collision

Before collision;
For car:mass=1.1×10^3, velocity=22
For truck:mass=2.3×10^3, velocity=0
After collision;
For car:mass=2.3×10^3, velocity=-11
For truck:mass=2.3×10^3, velocity=V
(1.1×10^3 × 22) + (2.3×10^3 × 0) = (1.1×10^3 × -11) + (2.3×10^3 × V)
24200 = -12100 + 2.3×10^3V
2.3×10^3V = 36300
V = 15.7 m/s
6 0
3 years ago
Read 2 more answers
Magnesium (average atomic mass = 24.3050 amu) consists of three isotopes with masses 23.9850 amu, 24.9858 amu, and 25.9826 amu.
iris [78.8K]

Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%

Explanation: Given that the average atomic mass(M) of magnesium

= 24.3050amu

Mass of first isotope (M1) = 23.9850amu

Mass of middle isotope (M2)=24.9858amu

Mass of last isotope(M3)= 25.9826amu

Total abundance = 1

Abundance of middle isotope = 0.10

Let abundance of first and last isotope be x and y respectively.

x+0.10+y =1

x = 0.90-y

M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope

24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y

Substitute x= 0.90-y

Then

y = 0.11

Since y=0.11, then

x= 0.90-0.11

x=0.79

Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%

5 0
3 years ago
50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?
d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3

Then you have to convert mL to L.

250mL*\frac{1L}{1000mL} = 0.250L

Now divide the mol by the L.

\frac{0.588mol NaNO3}{0.250L} = 2.352 M

Round to the smallest number of significant figures = 2.35M

7 0
3 years ago
Calculează masa zaharului si volumul apei necesare pentru prepararea 500g de soluție de zahăr cu partea de masa 20%
Sedbober [7]
T = 20 %  : 20 / 100 = 0.2

m1 = solute 

m2 = Solvent

T = m1 / m1 + m2

0.2 = 500 g / 500 g + m2

0.2 * ( 500 + m2 ) = 500

0.2 * 500 + 0.2 m2 = 500

100 + 0.2 m2 = 500

0.2 m2 = 500 - 100

0.2 m2 = 400

m2 = 400 / 0.2

m2 = 2000 g of water

hope this helps!



7 0
3 years ago
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