What is changed when matter undergoes a physical change?

What is the molarity of an hno3 solution if 51.0 ml is needed to react with 25.0 ml of 0.150 mkoh solution? the equation is hno3

artcher [175]

The Answer should be .074 M

6 0

2 years ago

How many moles of HBr are present in 250 ml of a 0.15 M HBr solution?

Fofino [41]

Answer:

0.0375 moles HBr

Explanation:

we are given;

Molarity of HBr solution as 0.15 M
Volume of the solution as 250 mL

We are required to determine the number of moles;

We need to know that;

Molarity = Moles ÷ Volume

Therefore;

Moles of HBr = Molarity of HBr solution × Volume of solution

Thus;

Moles of HBr = 0.15 M × 0.25 L

= 0.0375 Moles

Thus, the number of moles of HBr solute is 0.0375 moles

6 0

2 years ago

Acetylene, C2H2, burns according to the following reaction: C2H2 5O2 --> 4CO2 2H2O. Suppose 1.20 g of C2H2 is mixed with 3.50

OLEGan [10]

WWWWWWAAAAAAPPPPPP!!!!!

8 0

3 years ago

The formula for binary ionic compound formed between cesium and fluorine

OleMash [197]

Answer:

Cesium fluoride(CsF)

Explanation:

A binary compound is a compound that is composed of 2 distinct element. An ionic compound is composed of ions, usually one is a metal why the other is a non metal. One element gives out electron to form cation and the other receives electron to form anion in a binary compound.

Cesium is a group 1 element and it has one valence electron and it can easily donate this 1 electron to form a bond with other element. Group 1 element are generally very reactive. Cesium is a metal

Fluorine is in group 7 of the periodic table and is a non metal .Fluorine have 7 valency electron and requires 1 electron to form a stable octet.

When cesium and fluorine bond to form a binary compound cesium donate 1 electron and fluorine receives the 1 electron for both element to form a stable octet. The formula for the binary ionic compound of cesium and fluorine can therefore be expressed as Cesium fluoride(CsF)

7 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago