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vekshin1
2 years ago
14

The volume of a gas-filled balloon is 30.0 L at 313 K and 1520 mm Hg pressure. What would the volume be at standard temperature

and pressure (STP)?
Chemistry
1 answer:
pashok25 [27]2 years ago
4 0
313 k
step-by-step-explanation
You might be interested in
If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T
Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

         7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔      log([Conjugate base]/[Acid])  = 0

⇔     [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

                ⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

                ⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

    = 7.55 + log(0.003 / 0.009) = 7.07

8 0
2 years ago
A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c
nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0
2 years ago
What is the percent yield if 155 grams of calcium carbonate is treated with 250 grams of hydrichloric acid andb142 grams of calc
dlinn [17]

Answer:

375 grams

Explanation:

7 0
3 years ago
Read 2 more answers
Calcium carbide ( CaC2) reacts with water to produce acetylene (C2H2) as shown in the unbalanced reaction below: CaC2(s)+H2O(g)-
MaRussiya [10]
Answer:
46.3g H2O

Explanation:
start by balancing it: CaC2(s) + 2H2O(g) -> Ca(OH)2(s) + C2H2(g)

then use factor label method to solve

82.4g CaC2 x (1 mol CaC2/64.10g CaC2) x (2 mol H2O/1 mol CaC2) x (18.016g H2O/1 mol H20) = 46.3g H2O
4 0
2 years ago
How could you safely determine if a base is stronger than an acid? Compare the taste of the base and the acid. Use a conductivit
Gnoma [55]
Use blue litmus paper. This is an indicator that can safely determine whether it is a base or an acid by changing color in response to the substance. This color indicates whether it is an acid or a base. Refer to the pH scale to see if the substance is basic or acidic.
6 0
2 years ago
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