Match the causes with their effects

How many protons does this atom have?:____

Allisa [31]

Answer:

1.6

2.8

3.6

Explanation:

4 0

3 years ago

Read 2 more answers

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

<u>As we know</u>–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

3 years ago

the volume of gas in a container was originally 3.24L, while at standard pressure 1.00 atm. what will the volume be if the press

Simora [160]

Hello!

To solve this problem, we will use the Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

$P1*V1=P2*V2 \\ \\ V2= \frac{P1*V1}{P2}= \frac{3,24 L * 1 atm}{1,20 atm}= 2,7 L$

So, the final volume after increasing the pressure would be 2,7 L. That means that volume decreases when the pressure increases

Have a nice day!

8 0

3 years ago

II need this asap !!!!!!!

Xelga [282]

Answer:

1. B.10 mol/l

2. D.

3.A.

Explanation:

Totoo po yan ganyan po yong sakin

7 0

3 years ago

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

matrenka [14]

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

5 0

3 years ago