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3 years ago

11

Which has the lowest freezing point?

1 answer:

Volgvan3 years ago

5 0

Answer:

Mercury has the lowest freezing point for metals, but helium takes the winner for all substances.

Explanation:

You might be interested in

if the mass of a box was 400 grams, and the length is 2cm the width is 2cm and the height is 1cm. what is the density?

oksano4ka [1.4K]

Answer:

The density of the box is 6.25g/cm³

Explanation:

400 ÷ 4³ = 6.25

4 0

3 years ago

Which types of orbitals are found in the principal energy level n = 2?

AfilCa [17]

Remember this:

1) n is principal quantum number and represents the energy level.

2) l is the second quantum number and represent the type of orbital.

3) l can take values from 0 to n - 1

4) each number of l is associated with a type of orbital. This table shows the equivalence:

l number       type of orbital

0                    s

1                    p

2                    d

3                     f

With that, you can tell that n = 2 permits l = 0 and 1, which is orbitals s and p.

Therefore, the answer is the option D) s, p.

3 0

3 years ago

Read 2 more answers

) How many moles are there in 2.00 x 10^19 molecules of calcium phosphate?

Tamiku [17]

Answer:

$\huge 3.322 \times {10}^{ - 5} \: \: moles \\$

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{2.00 \times {10}^{19} }{6.02 \times {10}^{23} } \\ \\ = 3.322 \times {10}^{ - 5} \: \: moles$

Hope this helps you

4 0

3 years ago

Someone help this is so hard

tigry1 [53]

i need points

Explanation:

7 0

3 years ago

Read 2 more answers

You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol

oee [108]

Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

$0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol$

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

$6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH$

Step 4: Calculate the molarity of the acetic acid solution

$M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M$

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

$6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g$

Step 6: Calculate the percentage of acetic acid in the solution

$\frac{0.4186g}{25.00mL} \times 100\% = 1.674 \%(m/V)$

6 0

3 years ago

Read 2 more answers