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zhuklara [117]
3 years ago
13

Definition: Water that falls to the ground as rain

Chemistry
1 answer:
Ugo [173]3 years ago
6 0
Precipitation.
That’s the answer
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A mixture of gases containing 0.20 mol of SO2 and 0.20 mol of O2 in a 4.0 L flask reacts to form SO3. If the temperature is 25ºC
diamong [38]

Answer : The pressure in the flask after reaction complete is, 2.4 atm

Explanation :

To calculate the pressure in the flask after reaction is complete we are using ideal gas equation.

PV=n_TRT\\\\P=(n_1+n_2)\times \frac{RT}{V}

where,

P = final pressure in the flask = ?

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 25^oC=273+25=298K

V = volume = 4.0 L

n_1 = moles of SO_2 = 0.20 mol

n_2 = moles of O_2 = 0.20 mol

Now put all the given values in the above expression, we get:

P=(0.20+0.20)mol\times \frac{(0.0821L.atm/mol.K)\times (298K)}{4.0L}

P=2.4atm

Thus, the pressure in the flask after reaction complete is, 2.4 atm

5 0
3 years ago
What powdery mineral is hazardous when airborne? Mercury, Radon, asbestos or uranium
LekaFEV [45]

Answer: Uranium

Explanation:

3 0
3 years ago
Help! Giving brainliest!!!!
Nuetrik [128]
For the triplet AUG, I got methionine, which is a start codon. For GCU, I got alanine.
7 0
3 years ago
Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.
Inga [223]

Answer:

1.5g/cm³

Explanation:

density=mass÷volume

mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g

volume of the cube = 10×10×10 = 1000cm³

density= divide 1500g÷1000cm = 1.5g/cm³

<h2>Density= 1.5g/cm³</h2>

YOUR WELCOME!

4 0
3 years ago
A piece of copper (12.0 g) is heated to 100.0 °C. A piece of chromium (also 12.0 g) is chilled in an ice bath to 0 °C. The speci
Gennadij [26K]

Answer:

(a) Slightly greater than 20.0 °C

(b) T_F=293.46K=20.3^oC

Explanation:

Hello,

In this case, since we are talking about the equilibrium temperature that will be reached when the copper, chromium and water samples get in contact, the following equation is useful to describe such situation:

\Delta H_{water}+\Delta H_{copper}+\Delta H_{chromium}=0\\

Thus, in terms of masses, heat capacities and temperatures we consider the final temperature as the unknown:

m_{water}Cp_{water}(T_F-T_{water})+m_{copper}Cp_{copper}(T_F-T_{copper})+m_{chromium}Cp_{chromium}(T_F-T_{chromium})=0In such a way, by knowing that the heat capacities of copper and chromium are 0.386 and 0.45 J/(g°C) respectively, by solving for the equilibrium temperature one has:

T_F=\frac{m_{water}Cp_{water}T_{water}+m_{Cu}Cp_{Cu}T_{Cu}+m_{Cr}Cp_{Cr}T_{Cr}}{m_{water}Cp_{water}+m_{Cu}Cp_{Cu}+m_{Cr}Cp_{Cr}}

T_F=\frac{200.0g*4.184\frac{J}{g*K}* 293.15K+12.0g*0.386\frac{J}{g*K} *373.15K+12.0g*0.45\frac{J}{g*K}*273.15K}{200.0g*4.184\frac{J}{g*K}+12.0g*0.386\frac{J}{g*K} +12.0g*0.45\frac{J}{g*K}}\\\\T_F=293.46K=20.3^oC

Hence, the resulting temperature of water turns out slightly greater than 20.0 °C.

Best regards.

7 0
3 years ago
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