Hi there! Let's solve this problem shall we!
⠀Volume = 10g
Mass = 2 mL
In this specific problem, they are asking us to find the <u><em>density </em></u>of the object. So,<u><em> using the information given to us</em></u> (volume and mass), let's solve the problem!
Now, if you remember, D = M ÷ V
So, let's fill in the blanks!
D = Our unknown value
M = 2mL
V = 10g
Here is the filled out formula:
D = M ÷ V
D = 2mL ÷ 10g
D = 5 g/mL
*Make sure you put the units for your final solution!*
Talk to them and listen to each other. if they aren’t ready to talk, give them space. once both of you are ready, you can make up and forgive each other. don’t bother them by asking a lot of questions and forcing them to talk to you. and if they’re doing that, tell them you need time to think. just be sure to talk to them, listen, and understand. tell each other both sides of the stories. of course, different situations can require different solutions. so resolve it when it’s time :)
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The artificial fixation of nitrogen (N2) has enormous energy, environmental, and societal impact, the most important of which is the synthesis of ammonia (NH3) for fertilizers that help support nearly half of the world's population.
<h3>Artificial fixation of nitrogen</h3>
a) The equilibrium constant expression is Kp=PCH4 PH2 OP CO×PH 23.
(b) (i) As the pressure increases, the equilibrium will shift to the left so that less number of moles are produced.
(ii) For an exothermic reaction, with the increase in temperature, the equilibrium will shift in the backward direction.
(iii) When a catalyst is used, the equilibrium is not disturbed. The equilibrium is quickly attained
To learn more about equilibrium constant visit the link
brainly.com/question/10038290
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The units of ppm means parts per million. Also, It is equivalent to milligrams per liter. It is one way of expressing concentration of a substance. It u<span>sually used to describe the concentration of something in water or soil. We calculate the mass of CaCO3 as follows:
Mass = 75 mg/L (.050 L) = <span>3.75 mg CaCO3</span></span>
