Mitosis has 4 steps and meiosis has 5 steps so its 9 steps total here's a photo to help you if you need it
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Answer:</u></h2>
(These are not rounded to the correct decimal)
130.94 atm
13,266.6 kPa
99,571.4 mmHg
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Explanation:</u></h2>
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PV = nRT
V = 245L
P = ?
R = 0.08206 (atm) , 8.314 (kPa) , 62.4 (mmHg)
T = 273.15 + 27 = 300.15K
n = 1302.5 moles
How I found (n).
5.21kg x 1000g/1kg x 1 mole/4.0g = 1302.5 moles
Now, plug all the numbers into the equation.
Pressure in atm = (1302.5)(0.08206)(300.15) / 245 = 130.94 atm (not rounded to the correct decimal)
Pressure in kPa = (1302.5)(8.314)(300.15) / 245 = 13,266.6 kPa (not rounded to the correct decimal)
Pressure in mmHg = (1302.5)(62.4)(300.15) / 245 = 99,571.4 mmHg (not rounded to the correct decimal)
Answer:
Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)
Explanation:
Part 1. Volume of reactant
(a) Balanced chemical equation.

(b) Moles of CuCl₂

(c) Moles of Na₃PO₄
The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

(d) Volume of Na₃PO₄

Part 2. Net ionic equation
(a) Molecular equation

(b) Ionic equation
You write molecular formulas for the solids, and you write the soluble ionic substances as ions.
According to the solubility rules, metal phosphates are insoluble.
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
(c) Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
<u>6Na⁺(aq)</u> + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + <u>6Cl⁻(aq)</u> ⟶ Cu₃(PO₄)₂(s) + <u>6Na⁺(aq)</u> + <u>6Cl⁻(aq)</u>
The net ionic equation is
3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)
Answer: 253.8
Explanation:
The molar mass of iodine is 126.904. Multiply that by two and you get approximately 253.8 grams in two moles.
Magma that cools quickly forms one kind of igneous rock, and magma that cools slowly forms another kind. When magma rise from deep within the earth and explodes out of a volcano, it is called lava, and it cools quickly on the surface. Rock formed in this way is called extrusive igneous rock.