Which hybridization scheme occurs about phosphorus when nitrogen forms a triple bond?

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Answer 1

Answer: Jun 21, 2015 · 3 posts · 1 author

Most of the time when nitrogen forms a double bond it will be sp2 hybridised. permalink; embed; save.

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Round each answer to the number of significant figures in the parentheses.

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Answer:

87.100. 4.4000. 0.012000. 9000. 6,040,850. 630.00. 100.1000. 0.00024000

I need help!! ASAP please..

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Answer:

164 g

Explanation:

What is the empirical formula of a compound that is 64.3 % c, 7.2 % h, and 28.5 % o by mass?

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Asnwer : Empirical formula of a compound is : $C_(3)H_(4)O$

Given information : C = 64.3 % , H = 7.2 % , O = 28.5 %

Step 1 : Convert the given percentage (%) to grams.

Explanation : Let the total mass of the compound be 100 grams.

Mass of C = 64.3 g

$(100g)* ((64.3percent))/((100percent)) = 64.3g$

Mass of H = 7.2 g

$(100g)* ((7.2percent))/((100percent)) = 7.2g$

Mass of O = 28.5 g

$(100g)* ((28.5percent))/((100percent)) = 28.5g$

Step 2 : Convert the grams of each compound to moles.

$Moles = (Grams)/(Molar mass)$

Molar mass of C = 12.0g/mol

Molar mass of H = 1.0 g/mol

Molar mass of O = 16.0g/mol

$Moles of C = (64.3g)/(12.0(g)/(mol))$

Moles of C = 5.36 mol

$Moles of H = (7.2g)/(1.0(g)/(mol))$

Moles of H = 7.2 mol

$Moles of O = (28.5g)/(16.0(g)/(mol))$

Moles of O = 1.78 mol

Step 3 : Find the mole ratio of C , H and O

Mole ratio is calculated by dividing the mole values by the smallest value.

Mole of C = 5.36 mol , Mole of H = 7.2 mol , Mol of O = 1.78 mol

Out of the three mole values , mole value of O that is 1.78 mol is less , so we divide all the mole values by 1.78 mol.

$Mole of C = (5.36mol)/(1.78mol) = 3.0$

$Mole of H = (7.2mol)/(1.78mol) = 4.0$

$Mole of O = (1.78mol)/(1.78mol) = 1.0$

C : H : O = 3 : 4 : 1

So empirical formula of the compound is $C_(3)H_(4)O_(1)$ or $C_(3)H_(4)O$

What is the difference between a volume that is delivered and a volume that is contained?

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Answer:

When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.

A substance with a high vapor pressure at normal temperatures is often referred to as volatile. As the temperature of a liquid increases, the kinetic energy of its molecules also increases and as the kinetic energy of the molecules increases, the number of molecules transitioning into a vapor also increases, thereby increasing the vapor pressure.

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As the temperature of a liquid or stable will increase its vapor strain additionally will increase. Conversely, vapor strain decreases because the temperature decreases.

The better the vapor strain of a substance, the extra the awareness of the compound withinside the gaseous section and the extra the quantity of vaporization

. Liquids range substantially of their vapor pressures. substance with a excessive vapor strain at everyday temperatures is regularly called volatile. The strain exhibited through vapor gift above a liquid floor is referred to as vapor strain. As the temperature of a liquid will increase, the kinetic strength of its molecules additionally will increase.

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Answer:

The ratio of solution 1 to solution 2 is 24.20 to 100.00.

Explanation:

We will mix V₁ (L) of solution 1 with V₂ (L) of solution 2 to get the final solution.

So the mole concentration in the final solution is calculated as below, note that C₁ is the concentration of solution 1, and C₂ is the concentration of solution 2

$[M] = (V_(1) C_(1) +V_(2)C_(2))/(V_(1)+V_(2)) = \frac{4 V_(1) + 0.1 V_(2)}_{V_(1)+V_(2)}}=0.86$

Then we can calculate for the ratio

$(V_(1))/(V_(2))=(0.86-0.10)/(4.00-0.86) =(0.76)/(3.14)$ or $(24.20)/(100.00)$

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