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3 years ago

35mL of acid with an unknown concentration is titrated to completion using 63mL of 3.0 MNaOH.

Chemistry

1 answer:

siniylev [52]3 years ago

4 0

Answer:

5.4 M.

Explanation:

At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.

M of acid = ??? M, V of acid = 35.0 mL.

M of NaOH = 3.0 M, V of NaOH = 63.0 mL.

∴ M of acid = (MV)NaOH / (V)acid = (3.0 M)(63.0 mL)/(35.0 mL) = 5.4 M.

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Which of the following is a unit of length? O A. A liter O B. A kilogram C. A meter O D. A degree

sergeinik [125]

Answer:

C.) A meter

Explanation:

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3 years ago

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

3 years ago

Which particle is emitted when an atom of 85Kr spontaneously decays?

stiv31 [10]

A Beta particles is emitted when an atom of 85Kr spontaneously decays.

5 0

3 years ago

Read 2 more answers

How many grams of sodium Phosphate is needed to prepare 5.0 L of a 2.6M solution ?

Diano4ka-milaya [45]

Molarity is the ratio of the moles and the volume. The mass of 2.6 M sodium phosphate solution is 2131.22 gms.

Mass is the product of the moles and the molar mass of the substance. It is given as,

Mass = Moles × Molar mass

The moles from molar concentration is used to calculate mass as:

Mass = Molarity × volume × molar mass

= 2.6 × 5.0 × 163.94

= 2131.22 gms

Therefore, 2131.22 gms is the mass of sodium phosphate.

Learn more about mass here:

brainly.com/question/9829994

#SPJ1

6 0

2 years ago

According to the molecular orbital energy-level diagram below, which one of the following statements is not correct about NO, NO

Lubov Fominskaja [6]

None of the statement is not correct about NO, NO+, and NO−.

Explanation:

Those molecules whose all subshells are completely filled are diamagnetic.

Those molecules whose any sub-shell is incomplete is termed paramagnetic.

Let us check the magnetism in all three molecules based on molecular orbital theory and Lewis structure

NO= number of electrons 11

It form N=O having configuration 1s2,2s2,2p6,3s1, it has 1 unpaired electron so it is paramagnetic.

Now NO-, number of electrons 12 having configuration 1s2,2s2,2p6,3s2, no unpaired electron hence dimagnetic.

Now NO+, number of electrons is 10 having configuration 1s2,2s2,2p6,3s2, dimagnetic

Let us calculate the bond order of NO-

N0- have 12 electrons as 5 of nitrogen, 6 of oxygen and 1 of - charge on oxygen.

So, bond order = $\frac{8-4}{2}$

= 2 double bonds

bond order of NO+, Number of electrons is 10

bond order= $\frac{8-2}{2}$

bond order = 3

bond order of NO, number of electrons 11

bond order= $\frac{8-3}{2}$

bond order=2.5

Larger the bond order shorter the bond.

4 0

3 years ago