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stira [4]
3 years ago
5

Static, sliding and rolling are types of friction. Please select the best answer from the choices provided

Chemistry
1 answer:
GarryVolchara [31]3 years ago
6 0
The answer is true ^-^

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Estructura molecular del óxido nitroso​
lozanna [386]

Answer:

yes

Explanation:

3 0
3 years ago
If the gram-formula mass of substance X is 180 g/mol, determine the molarity of the solution at Point E
Alina [70]
<h3><u>Answer;</u></h3>

<u>= 5 M or 5 moles/liter</u>

<h3><u>Explanation;</u></h3>

At point E, 90 g of substances X are dissolved in 100 g of the solvent.

100g of the solvent is equal to 100 ml

Molarity is the number of moles of a substance in one liter of a solvent.

90 g of X are in 100 ml

But; the RFM of X = 180 g/l

Therefore; the moles of X in 90 g = 90/180

                                                        = 0.5 moles

Therefore;

0.5 moles of X are contained in 100 ml of the solvent;

Thus, molarity = 0.5 × 1000/100

                       =<u> 5 M or 5 moles/liter</u>

7 0
3 years ago
The box in the above picture is falling from the top of a building to the ground. Two major forces are acting on the box as it f
MAVERICK [17]

Answer:

I think it's B

Explanation:

apologies if I get this wrong

8 0
3 years ago
Read 2 more answers
Wastewater discharged into a stream by a sugar refinery contains 3.40 g of sucrose (C12H22O11) per liter. A government-industry
Ipatiy [6.2K]

<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm

<u>Explanation:</u>

To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

\pi=iMRT

or,

\pi=i\times \frac{m_{solute}}{M_{solute}\times V_{solution}\text{ (in L)}}}\times RT

where,

\pi = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

m_{solute} = mass of sucrose = 3.40 g

M_{solute} = molar mass of sucrose = 342.3 g/mol

V_{solution} = Volume of solution = 1 L

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 20^oC=[20+273]K=293K

Putting values in above equation, we get:

\pi =1\times \frac{3.40g}{342.3g/mol\times 1}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 293K\\\\\pi =0.239atm

Hence, the pressure that must be applied to the apparatus is 0.239 atm

3 0
3 years ago
Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen
Mkey [24]

Answer:

Oxygen is limiting reactant

Explanation:

2 H2  +   O2  ======> 2 H2 O

from this equation (and periodic table) you can see that

  4 gm of H combine with 32 gm O2  

     H / O  =  4/32 = 1/8

       32 /16    =  2/1    shows O is limiter

         for 32 gm H you will need 256 gm O   and you only have 16 gm

3 0
2 years ago
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