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melisa1 [442]
3 years ago
11

Help me please fast​

Chemistry
1 answer:
Rufina [12.5K]3 years ago
3 0

Answer:

He

Be

Ar are the noble gases

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How many molecules are in 3 moles of NaCl?
PtichkaEL [24]

Answer:

Given : No. Of moles = 1.5

To calculate : no. Of molecules =N

We know that moles = N / 6.022 x 10²³

Therefore, 1.5 x 6.022 x 10²³ = N

Hence N = 9.0330x 10²³ molecules

3 0
3 years ago
The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?
Umnica [9.8K]

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

3 0
3 years ago
Which process involves an increase in entropy? crystallization of a solute from a solution ice melting into liquid water iodine
GrogVix [38]
Ice melting into liquid water.
7 0
3 years ago
Read 2 more answers
If 25 mL of a HCl solution of unknown concentration was neutralized with 10 mL of a 0.30 M NaOH solution, what was the original
QveST [7]

Answer:

0.12 M

Explanation:

Step 1: Write the balanced equation

NaOH + HCl ⇒ NaCl + H₂O

Step 2: Calculate the reacting moles of NaOH

10 mL of a 0.30 M NaOH solution react.

0.010L \times \frac{0.30mol}{L} = 3.0 \times 10^{-3} mol

Step 3: Calculate the reacting moles of HCl

The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 1/1 × 3.0 × 10⁻³ mol = 3.0 × 10⁻³ mol.

Step 4: Calculate the concentration of HCl

3.0 × 10⁻³ mol of HCl are in 25 mL of solution.

M = \frac{3.0 \times 10^{-3} mol}{0.025L} = 0.12 M

5 0
3 years ago
50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

The equation of the reaction is;

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

We can use the titration formula;

CAVA/CBVB = NA/NB

CA= concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

CB = 0.010 M

VB = 50.0 ml

CA = 0.50 M

VA = ?

NA = 1

NB = 1

Substituting values;

CAVANB = CBVBNA

VA =  0.010 ×  50.0 × 1/ 0.50 × 1

VA = 1 ml

Since the total volume of acid used is 1 ml and each drop contains 0.040 ml

The number of drops required is 1ml/0.040 ml = 25 drops

Learn more: brainly.com/question/1527403

4 0
3 years ago
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