How many C-13 atoms are present, on average, in a 23000-atom sample of carbon?

The metals in Groups 1A, 2A, and 3A ____.

choli [55]

All of these metals from cations, so they lost electrons when forming ions. The answer is D.

3 0

3 years ago

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Which states of matter have particles that move independently of one another with very little attraction?

maks197457 [2]

Gas -- move very independently as compared to a liquid, for instance.

Plasma -- related to a gas although it has positively charged ions and negatively charged electrons.

8 0

3 years ago

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Please help me on this question

Goshia [24]

The answer to your question is 9.60.

8 0

2 years ago

A) A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M

Wewaii [24]

Answer:

i) 5.00 mL of 1.00 M NaOH: before the equivalence point

ii) 50.0 mL of 1.00 M NaOH: before the equivalence point

iii) 100 mL of 1.00 M NaOH: at the equivalence point

iv) 150 mL of 1.00 M NaOH: after the equivalence point

v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

$M=n/v\implies n=M\times V=1.00M\times 0.100L=0.100mol$

2. Now calculate the number of moles of NaOH for every condition (addition)

i) 5.00 mL of 1.00 M NaOH

$n=0.00500liter\times 1.00M=0.00500molNaOH$

Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

ii) 50.0 mL of 1.00 M NaOH

$n=0.0500liter\times 1.00M=0.0500molNaOH$

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

iii) 100 mL of 1.00 M NaOH

$n=0.100liter\times 1.00M=0.100molNaOH$

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

iv) 150 mL of 1.00 M NaOH

$n=0.150liter\times 1.00M=0.150molNaOH$

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

v) 200 mL of 1.00 M NaOH

This is more volume of NaOH, then this is also after the equivalence point.

5 0

3 years ago

Fe(ii) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts fe(ii)

strojnjashka [21]

0.0340 g O2

Step 1. Write the balanced chemical equation

4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]

= 4.250 mmol Fe^(2+)

Step 3. Calculate the moles of O2

Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]

= 1.062 mmol O2

Step 4. Calculate the mass of O2

Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2

= 0.0340 g O2

8 0

3 years ago