Which part of the circulatory system contains blood rich in carbon dioxide

What will increased temperature in a reaction cause

uranmaximum [27]

I think your answer is 2 because the temperature will rise so the particles will move faster

7 0

3 years ago

Write the formula for potassium oxide. why do you not need prefixes in the name

Molodets [167]

Potassium oxide: K₂O.

There's no need for prefixes since K₂O is an ionic compound.

<h3>Explanation</h3>

Find the two elements on a periodic table:

Potassium- K- on the left end of period four.
Oxygen- O- near the right end of periodic two.

Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.

Potassium is a metal,
Oxygen is a nonmetal.

A metal and a nonmetal combine to form an ionic compound. Potassium oxide is likely to be an ionic compound. It contains two types of ions:

Potassium ions: Potassium is group 1 of the periodic table. It is an alkaline metal. Like other alkaline metals such as sodium Na, potassium K tends to lose one electron and form ions of charge +1 in compounds. The ion would be K⁺.

Oxide ions from oxygen: Oxygen is the second most electronegative element on the periodic table. It tends to gain two electrons and form the oxide ion $\text{O}^{2-}$ when it combines with metals.

The two types of ions carry opposite charges. They shall pair up at a certain ratio such that they balance the charge on each other. The charge on each $\text{O}^{2-}$ ion is twice that on a $\text{K}^{+}$ ion. Each $\text{K}^{+}$ would pair up with two $\text{O}^{2-}$ . Hence the subscript in the formula: $\text{K}_{\bf 2}\text{O}$ .

There are two classes of compounds:

Covalent compounds, which need prefixes, and
Ionic compounds, which need no prefix.

Prefixes are needed only in covalent compounds. For instance in the covalent compound carbon dioxide $\text{CO}_2$ , the prefix di- indicates that there are two oxygen atoms in the formula $\text{CO}_2$ . However, there's no need for prefix in ionic compounds such as $\text{K}_2\text{O}$ .

7 0

3 years ago

The vapor above a mixture of pentane and hexane at room temperature contains 35.5% pentane by mass. What is the mass percent com

FrozenT [24]

The mass percent of Pentane in solution is 16.49%

The mass percent of Hexane in solution is 83.51%

Explanation:

Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.

Convert these values to mol% using their molecular weights:

Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol

Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol

Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%

Hexane mol%: yh = 100 - 39.68 = 60.32%

Pp-vap = 425 torr = 0.555atm

Ph-vap = 151 torr = 0.199atm

From Raoult's law we know:

Pp = xp $\times$ Pp - vap = yp $\times$ Pt (1)

Ph = xh $\times$ Ph - vap = yh $\times$ Pt (2)

Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:

(1 - xp) $\times$ Ph - vap = (1 - yp) $\times$ Pt (3)

Substituting (1) into (3) we get:

(1-xp) $\times$ Ph - vap = (1 - yp) $\times$ xp $\times$ Pp - vap / yp (4)

Rearrange for xp:

xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)

Subbing in the values we find:

Pentane mol% in solution: xp = 19.08%

Hexane mol% in solution: xh = 80.92%

Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:

mp = 0.1908 mol $\times$ 72.15 g/mol

= 13.766 g

mh = 0.8092 mol $\times$ 86.18 g/mol

= 69.737 g

Mass% of Pentane solution = 13.766/(13.766+69.737)

= 16.49%

Mass% of Hexane solution = 83.51%

8 0

3 years ago

Describe an experiment to determine the density of a liquid

Tju [1.3M]

Answer:

Take the measuring cylinder and measure its mass, in grams, as accurately as possible. Take the measuring cylinder off the balance and add the water carefully. Put the measuring cylinder back on the balance. Measure and record the new mass . Repeat the procedure, recording the volume and total mass, until the measuring cylinder is full. Then, for each volume calculate the mass of the liquid alone.Repeat steps 1 to 3 Draw a graph of mass of liquid against volume . Calculate the density of each liquid from the gradient of its graph line.

Explanation:

5 0

2 years ago

1. Methane (CH4) combusts as shown in the equation below. If 3 moles of methane were used in the combustion, what would be the c

umka2103 [35]

Answer:

The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ

Explanation:

Step 1: The balanced equation

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH = -802 kJ

Step 2: Given data

We notice that for 1 mole of methane (CH4), we need 2 moles of O2 to produce : 1 mole of CO2 and 2 moles of H20.

The enthalpy change of combustion, given here as Δ H , tells us how much heat is either absorbed or released by the combustion of one mole of a substance.

In this case: we notice that the combustion of 1 mole of methane gives off (because of the negative number), 802.3 kJ of heat.

Step 3: calculate the enthalpy change for 3 moles

The -802 kj is the enthalpy change for 1 mole

The change in enthalpy for 3 moles = 3* -802 kJ = -2406 kJ

The change in enthalpy in the combustion of 3 moles of methane = -2406 kJ

5 0

3 years ago