7.

Mrs tuckers class is doing an experiment they fill up a cup with ice water they measure the temperature of both the water and th

ELEN [110]

C.

Hope this is the correct answer

8 0

3 years ago

When a lithium atom forms an Li+ ion, the lithium atom

gizmo_the_mogwai [7]

4, loses an electron. a plus sign indicates an atom is losing an electron while a minus sign indicates an atom is gaining an electron
hope this helps!

3 0

3 years ago

Read 2 more answers

1 lb of CO2 occupies 0.6 ft^3 at a pressure of 200 psi. Determine the temperature of the system.

Anarel [89]

<u>Answer:</u> The temperature of the system is 273 K

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of carbon dioxide = 1 lb = 453.6 g (Conversion factor: 1 lb = 453.6 g)

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of carbon dioxide}=\frac{453.6g}{44g/mol}=10.31mol$

To calculate the temperature of gas, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of carbon dioxide = 200 psia = 13.6 atm (Conversion factor: 1 psia = 0.068 atm)

V = Volume of carbon dioxide = $0.6ft^3=16.992L$ (Conversion factor: $1ft^3=28.32L$ )

n = number of moles of carbon dioxide = 10.31 mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the system = ?

Putting values in above equation, we get:

$13.6atm\times 16.992L=10.31mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times T\\\\T=273K$

Hence, the temperature of the system is 273 K

8 0

4 years ago

⦁ Find the concentration of H+, OH-, PH and POH of 0.03 M of magnesium hydroxide which ionizes to the extent of only 1 /3 in aqu

lions [1.4K]

Answer:

$pH=12.3\\\\pOH=1.7\\$

$[H^+]=5x10^{-13}M$

$[OH^-]=0.02M$

Explanation:

Hello there!

In this case, according to the given ionization of magnesium hydroxide, it is possible for us to set up the following reaction:

$Mg(OH)_2(s)\rightleftharpoons Mg^{2+}(aq)+2OH^-(aq)$

Thus, since the ionization occurs at an extent of 1/3, we can set up the following relationship:

$\frac{1}{3} =\frac{x}{[Mg(OH)_2]}$

Thus, x for this problem is:

$x=\frac{[Mg(OH)_2]}{3}=\frac{0.03M}{3}\\\\x= 0.01M$

Now, according to an ICE table, we have that:

$[OH^-]=2x=2*0.01M=0.02M$

Therefore, we can calculate the H^+, pH and pOH now:

$[H^+]=\frac{1x10^{-14}}{0.02}=5x10^{-13}M$

$pH=-log(5x10^{-13})=12.3\\\\pOH=14-pH=14-12.3=1.7$

Best regards!

4 0

3 years ago

I need to find the reactants and products for each of these 7 problems. I need it in 3/5 hours help please!!!

Nezavi [6.7K]

Answer:

The arrow points from the reactants to the products, so just follow the arrows.

Explanation:

some have the reactants on the left and the products on the right, and others are the opposite... just know that

reactants---------> products

or

products<-----------reactants

4 0

4 years ago