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alina1380 [7]
2 years ago
12

Please please help me

Chemistry
1 answer:
ivolga24 [154]2 years ago
5 0
The answer is number four but the same time I don’t really know it’s like ha ha ha ha ha ha ha ha ha ha sorry
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The nickel(II) ion is commonly dissolved in solution using nickel(II) nitrate hexahydrate, Ni(NO3)2.6H2O. The nickel(II) ion pre
Juliette [100K]
Ni(OH)₂ ⇄ Ni⁺² + 2 OH⁻
Ksp = [Ni⁺²][OH⁻]²  = S (2S)² = 4S³
where S is molar solubility.
at pH = 10 
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴ 
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
S = \frac{6 x 10^{-16} }{10^{-8} } = 6 x 10⁻⁸ M
8 0
3 years ago
You are provided with a compound fertilizer, 40-15-10. Calculate the quantity of fertilizer to add to a one hectare field to sup
Kryger [21]

Answer:

The correct answer is a) 300 Kg b) 225 Kg c) 400 Kg and d) 600 Kg.

Explanation:

Based on the given value, that is. 40-15-10 shows the compositions of nitrogen, phosphorus, and potassium found in the compound fertilizer in the form of percentage.  

41 percent nitrogen is equal to 40/100 = 0.4, 15 percent phosphorus is equal to 15/100 = 0.15, and 10 percent potassium is equal to 10/100 = 0.1.  

To find the amount of fertilizer, which is needed per hectare to supply the different nutrients per hectare, there is a need to divide the quantity to be supplied by the percentage composition, now the values comes out as:  

a) At 120 Kg per ha, the required nitrogen will be 120/0.4 = 300 Kilogram of fertilizer.  

b) At 90 kg per ha, the required nitrogen will be 90/0.4 = 225 Kilograms of fertilizer.  

c) At 60 kg per ha, the required phosphorus will be 60/0.15 = 400 Kilograms of fertilizer.  

d) At 60 Kg per ha, the required potassium will be 60/0.1 = 600 Kilograms of fertilizer.  

5 0
2 years ago
This type of thermometer relies on a liquid to contract will colder and expand when warmed. A)infrared B)thermistor
Anarel [89]

Answer:  A

Explanation:

4 0
2 years ago
A food substance kept at 0°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6
ZanzabumX [31]

Answer:

1.   67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

Given that:

time t_1 = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time t_2 = 10.6 hours

Temperature T_1 = 0° C = (0+273 )K = 273 K

Temperature T_2 = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since (\frac{k_2}{k_1}=\frac{t_2}{t_1})

Then we can rewrite the above expression as:

In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})

In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})

2.934 = \frac{E_a}{8.314}(\frac{30}{82719})

2.934 = \frac{30E_a}{687725.766}

30E_a = 2.934 *687725.766

E_a = \frac{2.934 *687725.766}{30}

E_a =67255.58 \ J/mol

E_a =67.2 \ kJ/mol

7 0
2 years ago
12 +1<br><br> C<br><br>6<br><br><br>How many protons, electrons, and neutrons are there
hram777 [196]
6 protons, 6 neutrons, and 5 electrons
4 0
2 years ago
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