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3 years ago

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If a person looking at a poster sees green instead of yellow and doesn't see red at all, this person most likely has color blind

ness where _______ nerves fail to respond to light properly. A. blue-sensitive B. green-sensitive C. red-sensitive D. yellow-sensitive

2 answers:

Inga [223]3 years ago

5 0

<span>People with protanopia are unable to sense any ‘red’ light, people with deuteranopia do not sense ‘green’ light and people with tritanopia cannot sense ‘blue’ light. If a person perceives the color green, then the yellow sensitive nerves must work somewhat effectively since green is a combination of yellow and blue. Red-sensitive nerves are most likely not responding properly for this person. The answer is C.</span>

statuscvo [17]3 years ago

5 0

C. Red sensitive. just took the test.

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of

Yanka [14]

Answer:

Mass of carbon dioxide produced = 52.8 g

Explanation:

Given data:

Mass of carbon react = 14.4 g

Mass of oxygen = 56.5 g

Mass of oxygen left = 18.1 g

Mass of carbon dioxide produced = ?

Solution:

C + O₂ → CO₂

Number of moles of C:

Number of moles = mass/ molar mass

Number of moles = 14.4 g/ 12 g/mol

Number of moles = 1.2 mol

18.1 g of oxygen left it means carbon is limiting reactant.

Now we will compare the moles of C with CO₂.

C : CO₂

1 : 1

1.2 : 1.2

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 1.2 mol × 44 g/mol

Mass = 52.8 g

8 0

3 years ago

Which carboxylic acid is used to prepare the ester shown?

valentinak56 [21]

Answer:

B

Explanation:

The general equation for the reaction of a carboxylic acid with an alkanol to form an ester is shown below;

RCOOH + ROH ------> RCOOR + H2O

Hence; the reactant carboxylic acid can only be the compound (CH3)2-CH-CH2-COOH in accordance with the general reaction equation shown above.

Hence the reaction is;

(CH3)2-CH-CH2-COOH + CH3-CH2OH -------> CH3CH2 OCO-CH2-CH-(CH3)2

3 0

3 years ago

Calculate the ph of a 0.20 m solution of kcn at 25.0 ∘c. express the ph numerically using two decimal places

storchak [24]

The pH of the solution is basically the negative logarithm of the concentration of hydrogen ions or H+. In equation form, pH = -log[H+]. It could also be in terms of the concentration of hydroxide ions or OH- as pOH, where pOH = -log[OH-]. The sum of pH and pOH is 14. These are the important equations to know when it comes to equilibrium pH problems.

KCN is a basic salt coming from the reaction of a weak acid, HCN, and a strong base, KOH. In the hydrolysis of KCN, only the strong conjugate base (SCB) is involved. Since HCN is the weak acid, the SCB is CN-. The reaction would be

CN- + H2O ⇔ HCN + OH-

The important data is the equilibrium constant of acidity of the weak acid. Ka for HCN is 6.2×10^-10. Then, let's do the ICE(Initial-Change-Equilibrium) analysis.

CN- + H2O ⇔ HCN + OH-

I 0.2 m ∞ 0 0
C -x ∞ +x +x
-----------------------------------------------------------
E 0.2-x +x +x

The value x denotes the number of moles CN- reacted. There is no value for H2O because the solution is dilute such that H2O>>>CN-. Then, we apply the ratio:

$K_{H} = \frac{ K_{W} }{ K_{A} } = \frac{[HCN][OH-]}{[CN-]}$

where K,H is the equilibrium constant of hydrolysis and Kw is equilibrium constant for water solvation which is equal to 1×10^-14. Therefore,

$K_{H} = \frac{ 1x10^-14}{ 6.2x10^-10 } = \frac{[X][X]}{[0.2-X]}$

x = 0.001788 m
Since the value of OH- is also x, then OH-=0.001788 m. Consequently,

pOH = -log(0.001788) = 2.75
pH = 14 - pOH = 14 - 2.75
pH = 11.25

4 0

3 years ago

Read 2 more answers

A first-order decomposition reaction has a rate constant of 0.00440 yr−1. How long does it take for [reactant] to reach 12.5% of

mina [271]

Answer:

473 year

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

To reach 12.5% of reactant means that 0.125 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.125

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.125=e^{-0.00440\times t}$

t = 473 year

8 0

3 years ago

Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at

enyata [817]

Explanation:

(A) As we know that carbonic acid ( $H_{2}CO_{3}$ ) and Sodium bicarbonate ( $NaHCO_{3}$ ) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

pH = $pK_{a} + log(\frac{[Salt]}{[Acid]})$ ........... (1)

So mathematically, if [Salt] = [Acid] then $\frac{[Salt]}{[Acid]}$ = 1 .

And, $log (\frac{[Salt]}{[Acid]})$ = 0

Therefore, equation (1) gives us the following.

pH = $pK_{a}$ (when acid and salt are equal in concentration)

Hence, $pK_{a}$ of $H_{2}CO_{3}$ (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = $pK_{a}$ = 6.35.

(B) $[NaHCO_{3}]$ = 0.045 M and, $[H_{2}CO_{3}]$ = 0.45 M

Hence, pH = $6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])$

= $6.35 + log(\frac{0.045}{0.45})$

= 6.35 + (-1)

= 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C) $[NaHCO_{3}]$ = 0.45 M $[H_{2}CO_{3}]$

= 0.045 M

So, pH = $6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})$

= $6.35 + log(\frac{0.45}{0.045})$

= 6.35 + (+1)

= 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0

3 years ago