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pav-90 [236]
3 years ago
9

In the diagram shown, what is occurring at the section marked 4?

Chemistry
1 answer:
astra-53 [7]3 years ago
4 0
You need to show an image of the diagram or else nobody knows what you’re referring to.
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Which action would speed up a chemical reaction in aqueous solution?
Vedmedyk [2.9K]
Adding more powdered reactants
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How do the cells in your body get energy?
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What is the pressure in millimeters of mercury of 0.0130 molmol of helium gas with a volume of 210. mLmL at 55 ∘C∘C? (Hint: You
juin [17]

Answer : The pressure of the helium gas is, 1269.2 mmHg

Explanation :

To calculate the pressure of the gas we are using ideal gas equation:

PV=nRT

where,

P = Pressure of He gas = ?

V = Volume of He gas = 210. mL = 0.210 L    (1 L = 1000 mL)

n = number of moles He = 0.0130 mole

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

T = Temperature of He gas = 55^oC=273+55=328K

Putting values in above equation, we get:

P\times 0.210L=0.0130mole\times (0.0821L.atm/mol.K)\times 328K

P=1.67atm=1269.2mmHg

Conversion used : (1 atm = 760 mmHg)

Thus, the pressure of the helium gas is, 1269.2 mmHg

4 0
3 years ago
A chemist has 60ml of a 3M NaCl solution. He needs to dilute it so the Molarity is 1M NaCl. How much water does he need to add?
REY [17]

Answer:

80 ml the right answer

Explanation:

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>you</em><em> </em><em>dude</em><em> </em>

8 0
3 years ago
Use the molar heat capacity for aluminum from table 1 to calculate the amount of energy needed to raise the temperature of 260.5
Nimfa-mama [501]
Unfortunately, you failed to include the table 1 from which the molar heat capacity of aluminum could have been obtained. However, as a general rule, the heat needed to raise the temperature of a certain substance by certain degrees is calculated through the equation,
                            H = mcpdT
where H is heat, m is mass, cp is specific heat capacity, and dT is change in temperature. From a reliable source, cp for aluminum is equal to 0.215 cal/g°C. Substituting this to the equation,
                               H = (260.5 g)(0.215 cal/g°C)(125°C - 0)
                                        H = 7000.94 cal
6 0
3 years ago
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