The acceleration of a runner who accelerates from 0 m/s to 3 m/s in 12s

kolbaska11 [484]

Answer: jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

Explanation:

4 0

2 years ago

what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio

koban [17]

Answer:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

Explanation:

In this case, we can start with the formula of Platinum (II) Chloride. The cation is the atom at the left of the name (in this case $Pt^+^2$ ) and the anion is the atom at the right of the name (in this case $Cl^-$ ). With this in mind, the formula would be $PtCl_2$ .

Now, if we used metallic copper we have to put in the reaction only the copper atom symbol $Cu$ . So, we have as reagents:

$Cu~+~PtCl_2->$

The question now is: What would be the products? To answer this, we have to remember "single displacement reactions". With a general reaction:

$A~+~BC->AB~+~C$

With this in mind, the reaction would be:

$Cu~+~PtCl_2->Pt~+~CuCl_2$

I hope it helps!

5 0

2 years ago

Compare and contrast physical changes with chemical changes.

Alekssandra [29.7K]

There are several differences between a physical and chemical change in matter or substances. A physical change in a substance doesn't change what the substance is. In a chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.

3 0

2 years ago

How many grams are there in 2.3*10^24 atoms of silver nitrate ?

Paul [167]

410g Ag

2.3*10^24 atoms

1 molcule Ag- 6.02g*10^3

6 0

2 years ago

Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)

Reika [66]

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

For first isotope:

% = 50.69 %

Mass = 78.9183 amu

For second isotope:

% = 49.31 %

Mass = 80.9163 amu

Thus,

$Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu$

$Average\ atomic\ mass=40.0036+39.8998\ amu$

Average atomic mass = 79.9034 amu

4 0

2 years ago