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3 years ago

All matter and elements are made of extremely small particles called?

Chemistry

1 answer:

rewona [7]3 years ago

4 0

Tiny subatomic particles

You might be interested in

Can show some work for 4x = -12

timama [110]

4x = -12

First, divide both sides by 4. / Your problem should look like: $x = -\frac{12}{4}$
Second, since 4 goes into 3 to get 12, simplify it by 3. / Your problem should look like: $x = -3$

Answer: x = -3

6 0

3 years ago

Read 2 more answers

Determine the length of the object shown A. 97.8 mm B. 97.80 mm C. 97 mm D. 98 mm

jeka94

Answer:

A. 97.8 mm

Explanation:

There are 10 divisions between 9 cm (90 mm) and the end of the ruler (10 cm or 100 mm).

Each division equals 1 mm.

Each 5 mm mark has a longer tick, and the dashed red line is between the 97 mm and 98 mm marks.

You would normally estimate to the nearest tenth of a division. An estimate of 0.8 mm is reasonable.

The length of the object is 97.0 mm + 0.8 mm = 97.8 mm.

B is wrong. You can't possibly estimate to the nearest hundredth of a division.

C is wrong. The dashed red line slightly before the 98 mm mark.

D is wrong. If the dashed red line were exactly on the 98 mm mark, you would record the measurement as 98.0 mm. This indicates that you measured the object to zero tenths on either side of the mark.

6 0

2 years ago

Please help I really need it ill give brainliest

madam [21]

Answer:

Sorry, I'm trying to think but i cant explain..

8 0

2 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

pls this is my second time asking these questions, i'm sorry, but i really need help!!! 1. what happens to atoms when there is a

dusya [7]

Answer:

the atoms of the original substances gain, lose and even share their very own electrons. write the symbols of the elements that form the compound. Write down the valency, and lastly go over valencies. This is all I got so far. I hope this helps

Explanation:

8 0

2 years ago