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3 years ago

To create a program with block-based code, the programmer needs to drag the blocks of code into the _____.

Chemistry

1 answer:

padilas [110]3 years ago

8 0

Editor. That’s where you make the code

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_ g of potassium reacts with 16 g of oxygen to produce 94 g of potassium oxide

natali 33 [55]

Due to the law of conservation of mass, _g K + 16g O=94g KO.
So, 94-16=78
78 is your answer

5 0

4 years ago

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9. A box is pushed 1.5 m to the right in 5 s. What is the box’s average speed to the nearest hundredth of a m/s? *

poizon [28]

Answer:

s = 0.30 m/s

Explanation:

Given data:

Distance travel = 1.5 m

Time taken = 5 s

Average speed of box = ?

Solution:

s = d/t

s = speed

d = distance

t = time

by putting values,

s = 1.5 m/ 5 s

s = 0.30 m/s

7 0

3 years ago

Which of these following terms are made of matter and which or not made of matter: air,light,radio,magnetic field,car,radio wave

viva [34]

Matter: radio, car, flashlight, textbook, human

Not matter: everything besides the matter items

3 0

3 years ago

True or flas a series of repeated fission reactions caused by the release of neutrons in each reaction

Dafna11 [192]

I think the answer is true but I'm not very sure

4 0

3 years ago

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The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

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

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

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

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

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago