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s2008m [1.1K]
3 years ago
7

Please help me I will give a brainleist​

Chemistry
1 answer:
Korvikt [17]3 years ago
7 0
When the dew point temperature and air temperature are equal, the air is said to be saturated. Dew point temperature is NEVER GREATER than the air temperature. Therefore, if the air cools, moisture must be removed from the air and this is accomplished through condensation.
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1. Describe how SHAPE can change during a collision. Be specific with your evidence and add detail to your answer.
a_sh-v [17]

Answer:

Hey

of course, the damage of a collision depends upon how fast to objects are moving at each other and how strong they are. If you have two tanks moving at each other 2 miles per hour it will be very little damage and the ->shape<- will not change much, maybe a dint or two. But if two balloons filled with water are moving at each other 5000 mph they will completely evoporate in a burst of light, and their ->shape<- will change very much. This is how shape and motion are related.

Hope it helped

spiky bob your answerer

4 0
3 years ago
Wich of these equations are balanced? a) H2SO4+2AL---&gt;AL2(SO4) b)2KCL+Pb(NO3)2---&gt;2KNO3+PbCL2
Zolol [24]

The answer is b since on a it says H2 but on the right side there is no H idk if you forgot to put H there so im guessing b. Have a good day

4 0
3 years ago
Which of the following is the correct balanced equation for the reaction in which methane (CH4) burns in atmospheric oxygen (0₂)
Talja [164]

Answer: \text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}

Explanation:

The unbalanced equation is

\text{CH}_{4}+\text{O}_{2} \longrightarrow \text{CO}_{2}+\text{H}_{2}\text{O}

Balancing this equation, we get:

\boxed{\text{CH}_{4}+2\text{O}_{2} \longrightarrow \text{CO}_{2}+2\text{H}_{2}\text{O}}

8 0
2 years ago
The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14
Elena-2011 [213]

Answer:

Age of the atifact is 4.2\times 10^{2} years

Explanation:

  • For first -order radioactive decay- A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}
  • A_{t} represents activity of radioactive nuclide after t time, A_{0} represents initial activity of radioactive nuclide and t_{\frac{1}{2}} represents half-life
  • Here, A_{t}=19Bq, A_{0}=20Bq and t_{\frac{1}{2}}=5.73\times 10^{3}years

Plug-in all the given values in the above equation-

19=20\times (\frac{1}{2})^{\frac{t}{5.73\times 10^{3}}}

or, t=4.2\times 10^{2}

So, age of the atifact is 4.2\times 10^{2} years

6 0
3 years ago
How many atoms or moles of each element are on either side of the yield arrow? Is this equation balanced? 2Na+H2O---&gt; NaOH+2H
Anton [14]
4 NA+4H2O —> 4NaOH+2H2

I hope that helped
4 0
3 years ago
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