Most of the fresh water that is available for human use cannot be seen

Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at

MariettaO [177]

Answer:

P_2 =0.51 atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

$\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}$

$0.008389261745 \times 273 = 4.5P_2$

$P_2 =\dfrac{0.008389261745 \times 273 }{4.5}$

$P_2 =0.51 \ atm$

4 0

2 years ago

The National Environmental Policy Act was established in 1965. True or false

Mariana [72]

The answer to your question is false

5 0

3 years ago

Read 2 more answers

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago

Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.

-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to analyze the reagents. We have valine an amino acid, in this kind of compounds we have an amine group ( $NH_2$ ) and a carboxylic acid group ( $COOH$ ). Additionally, we have an alcohol ( $CH_3CH_2OH$ ) in the presence of HCl (a strong acid) in the first step, and a base ( $OH^-$ ).

When we have an acid and an alcohol in a vessel we will have an esterification reaction. In other words, an ester is produced. As the first step, the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the second step, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In step 3, a proton is transferred to produce a better leaving group ( $H_2O$ ). In step 4, a water molecule leaves the main structure to produce again the double bond C=O. Finally, a base ( $OH^-$ ) removes the hydrogen from the C=O bond to produce ethyl L-valinate