Answer:
D. all of these
Explanation:
In the classification of organisms, they are placed into groups which they appropriately fit into.
All the choices given correctly shows the process of classifying organisms. This aspect of biology is described as taxonomy.
- Based on characteristics of organisms, they are put into the same groups
- Also, we can identify the related family and genus in which a discovered organism can be put into.
- They phylogenetic relationship between organisms are important for classification.
Answer:
a) O²⁻ + H₂O <-----> OH⁻ + OH⁻
Acid = H₂O
Base = O²⁻
Conjugate Acid = OH⁻
Conjugate Base = OH⁻
b) HClO₄ + H₂SO₄ <-----> ClO₄⁻ + H₃SO₄⁺
Acid = HClO₄
Base = H₂SO₄
Conjugate Acid = H₃SO₄⁺
Conjugate Base = ClO₄⁻
c) NH₃ + HNO₃ <--> NH4⁺ + NO₃⁻
Acid = HNO₃
Base = NH₃
Conjugate Acid = NH4⁺
Conjugate Base = NO₃⁻
Explanation:
- Acids are molecules or ions capable of donating a proton (H⁺).
- Bases are molecules or ions that readily accept the H⁺ from acids.
- Conjugate Acids, according to the Brønsted–Lowry acid–base theory, are chemical compounds formed after the reception of a proton (H⁺) from an acid by a base.
- Conjugate Bases, Brønsted–Lowry acid–base theory, are the leftovers from when acids donate their proton (H⁺).
Answer:
Percent yield of PI3 = 95.4%
Explanation:
This is the reaction:
2P (s) + 3I2 (g) > 2PI3 (g)
Let's determine the moles of iodine that has reacted.
58.6 g / 253.8 g/mol = 0.231 mol
Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.
3 moles of I2 react to make 2 moles of PI3
0.231 moles of I2 would make (0.231 .2) / 3 = 0.154 moles of PI3
As we have produced 0.147 moles let's determine the percent yield.
(Yield produced / Theoretical yield) . 100 > (0.147 / 0.154) . 100 = 95.4%
Answer:
Explanation:
To determine the molar volume of the gas according to the equation at stp;
1 mole = 22.4 dm³
2 moles = 44.8 dm³
To determine the mass of NaN₃ inflated according to the equation
2NaN₃ (where Na = 23g and N = 14) = (2 × 23) + 2(14 × 3)
= 130 g
Hence, if 130g of NaN₃ is required to inflate 44.8 dm³ airbag upon impact
what mass of NaN₃ is required to fully inflate the air bag upon impact;
130g ⇒ 44.8 dm³
? ⇒ 11.9 dm³ (dm³ is same as L)
? = 130 × 11.9/44.8
? = 34.5g
34.5g of NaN₃ is required to fully inflate 11.9 L of air bag upon impact
