Elements in the same group (family) appeared to have the same what?

Enter your answer in the provided box.

nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, $V_1=43.1\ L$

Initial temperature, $T_1=24^{\circ} C=24+273=297\ K$

Final temperature, $T_2=78^{\circ} C=78+273=351\ K$

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L$

So, the final volume of the gas is 50.93 L.

4 0

2 years ago

Think about chemicals you may have in your home and list as many as you can think of:

MAVERICK [17]

Sucrose; C12H22O11

C9H8O4; acetyl salicylic acid

H2O2; hydrogen peroxide,

NaOH; sodium hydroxideExplanation:

6 0

2 years ago

Read 2 more answers

KHP is a monoprotic acid which provides one H+ ion. How would your results be affected if a diprotic acid (such as sulfuric acid

kari74 [83]

Explanation:

A.

In a diprotic acid, 2 moles of H+ ions is released. Therefore, number of moles of H+ in a diprotic acid = 2 × number of moles of H+ of monoprotic acid.

B.

Equation of the reaction

2NaOH + H2SO4 --> Na2SO4 + 2H2O

Number of moles of H2SO4 = molar concentration × volume

= 0.75 × 0.0105

= 0.007875 moles.

By stoichiometry, since 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, number of moles of NaOH = 2 × 0.007875

= 0.01575 moles.

Molar concentration of NaOH = number of moles ÷ volume

= 0.01575 ÷ 0.0175

= 0.9 M of NaOH.

7 0

2 years ago

At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i

olganol [36]

Answer:

$K_c=0.0867$

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.760}{1.50\ L}$

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

$\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}$