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2 years ago

Arrange these Alkaline Earth metals from lowest to highest electronegativity: Ca, Be, Ba, Sr

Chemistry

1 answer:

a_sh-v [17]2 years ago

6 0

Ca is -2

Be is stabilized

Ba is -4

Sr is -3

so it would be => Ba, Sr, Ca, Be

have a wonderful day! :)

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Na + Cl --> NaCl

dalvyx [7]

Answer:

38.152 g NaCl would be produced.

Explanation:

$\frac{(23 + 35.5) \times 15}{23} \\ = 38.152 \: \: g$

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2 years ago

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How many liters in 4.4 grams of CO2 at STP?

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2.24 Liters are in 4.4 grams of CO2 at STP

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Molarity of a salt water solution of "0.47" moles of NaCl dissolved in a volume of 0.25L

asambeis [7]

Answer:

1.88 M

Explanation:

The following data were obtained from the question:

Mole of NaCl = 0.47 mole

Volume of solution = 0.25L

Molarity =?

Molarity is defined as the mole of solute per unit litre of the solution. It can represented mathematically as:

Molarity = mole /Volume

Using the above formula, the molarity of the salt water solution can be obtained as follow:

Molarity = 0.47/0.25

Molarity = 1.88 M

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2 years ago

50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

= 25.00 x 0.200

= 5.00 m-mol

= 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

= 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

= 0.167 M

Now the ICE table :

$NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M) 0.167 0 0

C (M) -x +x +x

E (M) 0.167-x x x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$ $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

= $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

= $- \log(1.9054 \times 10^{-6} \ M)$

=5.72

Now, since pH + pOH = 14

pH = 14.00 - 5.72

= 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0

3 years ago