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3 years ago

13

During lab, you learned about exothermic and endothermic heats of solution. which compounds below could be used for a hot pack?

(select all that apply.)

1 answer:

tatiyna3 years ago

8 0

What are the options? It says select all that apply...

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List the following bond types in order of increasing strength: non-polar covalent bonds, ionic bonds, hydrogen bonds, polar cova

ss7ja [257]

Answer: Option (B) is the correct answer.

Explanation:

An ionic bond is formed by the sharing of electrons between two chemically combining atoms.

In an ionic bond, there occurs attraction between oppositely charged ions due to which there occurs strong forces of attraction between them. Therefore, ionic bonds are the strongest bonds.

A polar covalent bond is formed due to unequal sharing of electrons between the combining atoms.

For example, $H_{2}O$ is a polar covalent compound. Partial opposite charges tend to develop on the atoms of a polar covalent compound.

A non-polar covalent bond is formed due to equal sharing of electrons between the combining atoms.

For example, $Cl_{2}$ is a non-polar covalent molecule. No partial charges will be there on the atoms of a non-polar covalent molecule.

A hydrogen bond is defined as the bond formed between a hydrogen atom and an electronegative atom.

For example, in HCl compound there occurs hydrogen bonding.

In this type of bond, dipole-dipole attractive interactions tend to take place. And, strength of hydrogen bonds is very weak.

Thus, we can conclude that given bond types are arranged in order of increasing strength as follows.

Hydrogen bonds < non-polar covalent bonds < polar covalent bonds < ionic bonds

8 0

3 years ago

1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L

anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol $(P^o_1)$ = 20.9 torr

Vapor presume of 2‑Propanol $(P^o_2)$ = 45.2 torr

Mole fraction of 1‑Propanol $(x_1)$ = 0.540

Mole fraction of 2‑Propanol $(x_2)$ = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

$p_1=x_1\times p^o_1$

where,

$p_1$ = partial vapor pressure of 1‑Propanol

$p^o_1$ = vapor pressure of pure substance 1‑Propanol

$x_1$ = mole fraction of 1‑Propanol

$p_1=(0.540)\times (20.9torr)=11.3torr$

and,

$p_2=x_2\times p^o_2$

where,

$p_2$ = partial vapor pressure of 2‑Propanol

$p^o_2$ = vapor pressure of pure substance 2‑Propanol

$x_2$ = mole fraction of 2‑Propanol

$p_2=(0.46)\times (45.2torr)=20.8torr$

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

$\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352$

and,

$\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648$

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0

3 years ago

What does percent composition tell you about a substance?

beks73 [17]

Answer:

Percent composition tells you which types of atoms (elements) are present in a molecule and their levels. Percent composition can also tell you about the different elements present in an ionic compound as well.

8 0

2 years ago

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

3 years ago

For the reaction IO3–(aq) + 5 I–(aq) + 6 H+(aq) → 3 I2(aq) + 3 H2O(l) the rate of disappearance of I–(aq) at a particular time a

scoray [572]

Answer: The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$IO_3^-(aq)+5I^-(aq)+6H^+(aq)\rightarrow 3I_2(aq)+3H_2O(l)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[I^-]}{5dt}=+\frac{d[I_2]}{3dt}$

Given: $-\frac{d[I^-]}{dt}]$ = $2.4\times 10^{-3}mol/Ls$

$+\frac{d[I_2]}{dt}=-\frac{3d[I^-]}{5dt}=-\frac{3}{5}\times 2.4\times 10^{-3}mol/Ls=1.44\times 10^{-3}mol/Ls$

The rate of appearance of $I_2(aq)$ is $1.44\times 10^{-3}mol/Ls$

6 0

3 years ago