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4 years ago

11

Imagine that a brown horse and a white horse cross to produce an offspring whose coat is made up of brown hairs and some white h

airs. Which part of dominance is this apart of

1 answer:

wlad13 [49]4 years ago

4 0

This is known as incomplete dominance, where the two traits are mixed together when both of the alleles are present.

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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of

Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0

3 years ago

3. What is the mass of 5 moles of Hydrogen sulfate?I

solong [7]

Answer:

490

98 for 1 mole, Hence for 5 moles 5 X 98 =490.

Explanation:

Brainliest please?

3 0

3 years ago

How many kJ of heat are released by the reaction of 25.0 g of Na2O2(s) in the following reaction? (M = 78.0 g/mol for Na2O2)

solong [7]

-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.

Explanation:

Given:

mass of Na2O2 = 25 grams

atomic mass of Na2O2 = 78 gram/mole

number of mole = $\frac{mass}{atomic mass of 1 mole}$

= $\frac{25}{78}$

=0. 32 moles

The balanced equation for the reaction:

2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ

It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.

similarly 0.3 moles of Na2O2 on reaction would give:

$\frac{126}{2}$ = $\frac{x}{0.32}$

x = $\frac{126 x 0.32}{2}$

= -20.16 KJ

Thus, - 20.16 KJ of energy will be released.

6 0

4 years ago

How many simple distillation columns are required to purify a stream containing five components into ﬁve 'pure"products? Sketch

katen-ka-za [31]

Answer: one simple distillation column is required to separate the stream into five pure products. With four different flat bottom flask, for collection of the distilled products

Explanation: simple distillation works with the difference in boiling points of the liquid to be separated. For the separation of five different constituent to be possible, we have to know the boiling points of the constituents.

For your understanding, let's define constituents in the liquid to be A, B, C, D, E. And the boiling points increases respectively. Start by heating the liquid to the boiling point of A to extract A. After a while check if the constituents A is still dropping in the flat bottom flask, if it has stopped dropping, it simply means that we have extracted all A constituents in the liquid, label the Flask A. Get another flask to extract constituent B.

Heat the mixture to the boiling point of B, after a while check if constituent B is still dropping in the flat bottom flask, if it has stopped dropping,it means that we have extracted all B constituent in the liquid, label the Flask B. Get another flask for C.

Repeat the same process for C and D.

After Extracting D we don't need to distillate E because we already have a pure form of E inside to the conical flask.

SEE PICTURE TO UNDERSTAND WHAT A SIMPLE DISTILLATION LOOKS LIKE

6 0

3 years ago

Hydrogen gas (H2) reacts with oxygen gas (O2) and produces water. If one mol of hydrogen reacts with one mole of oxygen, what is

Amiraneli [1.4K]

HI

So, the formula for water is H2O

When you have the same amount of the reactants , hydrogen will be the limiting reactant.
Limiting reactant is the thing that runs out first.

3 0

3 years ago