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2 years ago

Which statement is true about SO4 2− ?

Chemistry

2 answers:

harkovskaia [24]2 years ago

6 0

Answer:

The answer to your question is: the third option

Explanation:

SO₄ ⁻²

Polyatomic ion is a molecule composed by 2 or more atoms, so SO₄⁻² is a polyatomic ion.

Cation: is an ion that has a positive charge.

Anion: is an ion that has a negative charge.

so SO₄ ⁻² is a polyatomic anion because it is composed by two elements and has a negative charge.

It is a polyatomic anion because it has more than one atom and a negative charge.

Nonamiya [84]2 years ago

4 0

It is a poly atomic anion because it has more than one atom and a negative charge

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A 25.00 mL sample of 0.320 M KOH is titrated with 0.750 M HBr at 25 °C.

Fudgin [204]

Answer:

a. pH = 13.50

b. pH = 13.15

Explanation:

Hello!

In this case, since the undergoing chemical reaction between KOH and HBr is:

$HBr+KOH\rightarrow KBr+H_2O$

As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:

$pOH=-log([OH^-])=-log(0.320)=0.50$

Thus, the pH is:

$pH+pOH=14\\pH=14-pOH=14-0.50\\pH=13.50$

Which is the same answer for a and b as they ask the same.

Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:

$n_{KOH}=0.02500mL*0.320mol/L=0.00800mol\\\\n_{HBr}=0.005L*0.750mol/L=0.00375mol$

It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:

$n_{KOH}^{remaining}=0.00425mol$

Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):

$[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M$

So the pOH and the pH turn out:

$pOH=-log(0.142)=0.849\\pH+pOH=14\\pH=14-pOH=14-0.849\\pH=13.15$

Best regards!

7 0

2 years ago

Sodium chloride or table salt is a compound formed when sodium loses its valence electron to chlorine. Which type of bond format

avanturin [10]

When Sodium chloride or table salt is a compound formed when sodium loses its valence electron to chlorine, the type of bond formation that takes place in sodium chloride is the Lewis Bond.

8 0

2 years ago

Read 2 more answers

When 35.47 g of sodium hydroxide react with boric acid (H3BO3), how many moles of sodium borate will be produced?

belka [17]

Answer:

5.83 g

Explanation:

First, you must start with a balanced equation so you can see the mole ratios.

NaOH + H₃BO₃ --> NaBO₂ + 2H₂O

You can see that it takes 1 mole of sodium hydroxide to form 1 mole of sodium borate. 1:1 ratio

Now you must calculate how many moles of NaOH 35.47 g equals.

Na = 22.99 amu

O = 15.99 amu

H = 1.008 amu

NaOH = 39.997 amu

35.47 g ÷ 39.997 amu = 0.08868 moles of NaOH

Since it's a 1:1 ratio, the same number of moles of NaBO₂ is created. Now you must convert moles to grams.

Na = 22.9 amu

B = 10.81 amu

2 O = 31.998 amu

NaBO₂ = 65.798 amu

0.08868 moles x 65.798 = 5.83 g

4 0

2 years ago

does the nitro group on the pyridine ring make the ring more electron rich or more electron deficient

Arte-miy333 [17]

Answer:

more electron deficient

Explanation:

The nitro group is an electron withdrawing group. It withdraws electrons from the pyridine ring by resonance.

This electron withdrawal by resonance makes the pyridine ring less electron rich or more electron deficient.

Hence, the nitro group makes the pyrinde ring more electron deficient

3 0

2 years ago

1. Mg + N2 —> MgN2

quester [9]

(1) IS THE BALANCED EQUATION AND REACTANT-> MG AND N2 AND PRODUCT ->MGN2.

(2) IS NOT BALANCED AND REACTANT->CF4 AND BR AND PRODUCT IS CBR4 AND F2

8 0

3 years ago