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GenaCL600 [577]

3 years ago

5

Why is butanone achiral?

1 answer:

maw [93]3 years ago

4 0

Because it has no <span> stereogenic carbon centres.</span>

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What is the empirical formula for C2H4S2?

egoroff_w [7]

Answer:

46

Explanation: ihjk

7 0

3 years ago

Calculate the ratio of effusion rates of cl2 to f2 .

WINSTONCH [101]

<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>

6 0

3 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

3 years ago

Help me please ❤❤❤

musickatia [10]

Answ burh i dont even know

Explanation: i dont even know this

7 0

3 years ago

What happens to a line when the y-intercept is changed? Check all that apply.

tamaranim1 [39]

Answer:

As the y-intercept increases, the graph of the line shifts up;

As the y-intercept decreases, the graph of the line shifts down

Explanation:

There are two ways to think about this problem. The first way would be the graphical approach:

if we only change the y-intercept, this means we keep the same slope;
y-axis is the vertical axis;
if we change the point at which the line crosses the y-axis, we either shift it upward for a higher y-intercept or downward for a lower y-intercept.

Now, thinking algebraically, a line has the following equation in a general form:

$y = mx + b$

The y-intercept is essentially obtained when x = 0, then:

y = b:

if we increase b value, the y value increases, so the graph shifts upward;
if we decrease b value, the y value decreases, so the graph shifts downward.

5 0

3 years ago