All categories

4 years ago

Wich of these is not a sign of chemical change 1 A gas is given off

Physics

1 answer:

mash [69]4 years ago

5 0

<h3>Answer:</h3>

2. The material dissolves

<h3>Explanation:</h3>

The Question tests the understanding of chemical change and whether we know the signs that show its occurrence.

First we need to know what is chemical change;

Chemical change is a type of change that involves formation of a new substance different from the original substance.

Second we need to identify the signs that indicate the occurrence of a chemical change

Chemical change in chemical reactions is characterized by various signs which include;

Production of bubbles
Color change
Formation of a precipitate
Change of odor
Change in temperature.

Therefore, in this case, the material dissolving is not a sign that chemical change has taken place.

You might be interested in

Fluids can flow and change shape to fit their containers. Which two states are<br> fluids?

mr_godi [17]

Answer: the two states that are fluid are;-

<u>#{1} liquid</u>

Explanation:

as we know that there are mainly three states of substance

but among them only two of them can fluid and takes the shape of the container that are liquid and gas

4 0

3 years ago

Four metals were kept in contact with each other. The table below shows the initial temperature of the metals.

disa [49]

Answer:

metal B

temperature flow from high region to low . metal B has more temperature than others

5 0

2 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

3 years ago

The batteries in the figure above are connected in

Elena-2011 [213]

The answer is a, series circuit.

5 0

4 years ago

Read 2 more answers

What happened in my room<br>

Natalija [7]

Answer:

<h2>A GHOST CAME! Booooo!!!!!!</h2>

Hah lol

5 0

3 years ago