Please answer all the questions

The rate constant for this first‑order reaction is 0.740 s − 1 0.740 s−1 at 400 ∘ C. 400 ∘C. A ⟶ products A⟶products How long, i

sesenic [268]

Answer:

1.668 S

Step-by-step explanation:

We are given that

$A\rightarrow$ Product

For first order reaction,

$t=\frac{1}{k}ln\frac{A_0}{A}$

$t=\frac{2.303}{k}log\frac{A_0}{A}$

Rate constant=k= $0.740s^{-1}$

Temperature= $T=400^{\circ}$

We have to find the time taken by the reaction when concentration of A to decrease from 0.790 M to 0.230 M.

$A_0=0.79 M$

$A=0.230$

Substitute the values then we get

$t=\frac{2.303}{0.740}log\frac{0.79}{0.23}$

$t=1.668 s$

Hence, it would take 1.668 s for the concentration of A to decrease from 0.790 M to 0.230 M.

4 0

4 years ago

Two skaters are racing toward a finish line of a race. The first skater has a 40 meter lead and is traveling at a rate of 12 met

Varvara68 [4.7K]

Answer:

20 seconds

Step-by-step explanation:

speed = distance/time

distance = speed * time

The faster racer will catch up to the slower racer in x meters ahead of the slower racer.

The faster racer is 40 meters behind the slower racer, so he travels x + 40 until he catches up.

Slower racer:

speed = 12 m/s

distance = x

time = t

Faster racer:

speed = 14 m/s

distance = x + 40

time = t (the same time as the slower racer)

Slower racer:

distance = speed * time

x = 12t

Faster racer:

x + 40 = 14t

System of equations:

x = 12t

x + 40 = 14t

Since the fist equation is already solved for x, substitute 12t for x in the second equation and solve for t.

12t + 40 = 14t

40 = 2t

20 = t

t = 20

The faster racer catches up in 20 seconds.

7 0

4 years ago

This is something about absolute value but i have no idea how to do it

Advocard [28]

You're looking for an inequality of the form $|t-b|\le a$ , or $-a\le t-b\le a$ , or $b-a\le t\le b+a$ .

Start with

$350^\circ\le t\le400^\circ$

Notice that 375º is the midpoint of this interval of temperatures. If we subtract all sides by 375º, then we get

$-25^\circ\le t-375^\circ\le25^\circ$

so we can set $a=25^\circ$ and $b=375^\circ$ , and we can rewrite this inequality in terms of absolute values as

$|t-375^\circ|\le25^\circ$

3 0

4 years ago

The temperature falls from 3°C to -4°C. What is the difference in these temperatures?

raketka [301]

Answer:

-1

Step-by-step explanation:

3- 4 = -1

3 0

4 years ago

Read 2 more answers

Solve the following linear quadratic system of equations algebraically. y=^2+3x-2 y+3=5x

USPshnik [31]

Answer:

$x=1$

$y=2$

Step-by-step explanation:

$y=x^2+3x-2$ , $y+3=5x$

Replace all occurrences of $y$ in $y+3=5x$ with $x^2+3x-2.$

$(x^2+3x-2)+3=5x$

$y=x^2+3x-2$

Add $-2$ and 3.

$x^2+3x+1=5x$

$y=x^2+3x-2$

Subtract 5x from both sides of the equation.

$x^2+3x+1-5x=0$

$y=x^2+3x-2$

Subtract 5x from 3x.

$x^2-2x+1=0$

$y=x^2+3x-2$

Rewrite 1 as $1^2$ .

$x^2-2x+1^2=0$

$y=x^2+3x-2$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2x=2$ · $x$ · $1$

$y=x^2+3x-2$

Rewrite the polynomial.

$x^2-2$ · $x$ · $1$ $+$ $1^2=0$

$y=x^2+3x-2$

Factor using the perfect square

trinomial rule $a^2-2ab+b^2=(a-b)^2,$

where a = x and b = 1.

$(x-1)^2=0$

$y=x^2+3x-2$

Set the $x-1$ equal to 0.

$x-1=0$

$y=x^2+3x-2$

Add 1 to both sides of the equation.

$x=1$

$y=x^2+3x-2$

Replace all occurrences of $x$ in

$y=x^2+3x-2$ with 1.

$y=(1)^2+3(1)-2$

$x=1$

$y=2$

3 0

3 years ago