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4 years ago

Which substance is a primary source of hydrocarbon

Chemistry

1 answer:

IceJOKER [234]4 years ago

6 0

Answer:

Hydrocarbons are molecules consisting of both hydrogen and carbon. They are most famous for being the primary constituent of fossil fuels, namely natural gas, petroleum, and coal. For this reason, fossil fuel resources are often referred to as hydrocarbon resources.Jun 25, 2018

https://energyeducation.ca ›

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Isotonic saline solution, which has the same osmotic pressure as blood, can be prepared by dissolving 0.923 g of nacl in enough

vova2212 [387]

<span>osmotic pressure = CRT to get C, get the number of moles of NaCl then divide it with volume of the solution mole NaCl = 0.923g / 58.442 g/mol mole NaCl = 0.0158 mol C = 0.0158 mol / 0.1L C = 0.158 mol/L So, osmotic pressure = 0.158 mol/L*(0.082058 L-atm/mol-K)*(25+273K) osmotic pressure = 3.86 atm</span>

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4 years ago

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3 years ago

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Un catión trivalente contiene 75 nucleones y 45 neutrones, Señale el número de electrones que posee dicho catión.

FromTheMoon [43]

Answer:

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Explanation:

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3 years ago

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

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4 years ago

Hey please come on help I needed this help

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