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4 years ago

An element's _____ is its row in the periodic table

Chemistry

2 answers:

Dmitry_Shevchenko [17]4 years ago

6 0

The answer is family

-Dominant- [34]4 years ago

3 0

Family is the answer.

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Hello! this is to my other post!

spin [16.1K]

I would pick the shrimp because it has the most chromosomes so that is more protein we when you eat shrimp you are getting the most protein out of all of them

Explanation: put this in your own words please

5 0

3 years ago

The mass of an electron is about 9.11 × 10 -9 kg., write this whole number

Pepsi [2]

Answer:0.00000009

Explanation:

9.11 × （1/1000000000）=0.00000009

8 0

3 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

The formula can be written for the calculation of moles as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

1 mole of AgNO₃ reacts with 1 mole of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

Thus, the concentration of NaCl = 0.3374 M

6 0

3 years ago

I need to know the weighted average please!

maria [59]

Answer:

63.5456 amu

Explanation:

$0.6917*62.9296+0.3083*64.9278= 63.5456\ amu$

8 0

3 years ago

buffer consists of 0.50 M NaH2PO4 and 0.40 M Na2HPO4. Phosphoric acid is a triprotic acid (K a 1 = 7.2 × 10 − 3, K a 2 = 6.3 × 1

Levart [38]

Answer:

The Ka value used for the buffer system is :- Ka2 = $6.3\times 10^{-8}$ pKa2 = 7.2

pH = 7.1

Explanation:

The buffer system given in the question is :-

$NaH_2PO_4$ and $Na_2HPO_4$

The reaction taking place is:-

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}$

The Ka value used for the buffer system is :- Ka2 = $6.3\times 10^{-8}$ pKa2 = 7.2

The pH can be calculated as:-

$pH=pKa+\log\frac{Na_2HPO_4}{NaH_2PO_4}$

pH = 7.2 + log 0.40/0.50 = 7.1

7 0

4 years ago