Which list of elements consists of a metal, a metalloid, and a nonmetal?

Consider the titration of sulfuric acid with sodium hydroxide. What volume (mL) of a 2.658M NaOH solution is required to fully t

KatRina [158]

The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

From the question;

Concentration of acid CA = 0.426M

Concentration of base CB = 2.658M

Volume of acid VA = 10.00mL

Volume of base VB = ?

Number of moles of acid NA = 1

Number of moles of base NB = 2

Using the relation;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB = 0.426M × 10.00mL × 2/ 2.658M × 1

VB = 3.2 mL

Learn more: brainly.com/question/6111443

8 0

2 years ago

Calculate in gramm the mass of 0.1 mole of hydrochloric acid(h=1 ,cl=35.5)

babunello [35]

Answer:

mass of HCl = 3.65 g

Explanation:

Data Given:

Moles of hydrochloric acid HCl = 0.1 mole

Mass in grams of hydrochloric acid HCl = ?

Solution:

Mole Formula

no. of moles = Mass in grams / molar mass

To find Mass in grams rearrange the above Formula

Mass in grams = no. of moles x molar mass . . . . . . . (1)

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Put values in equation 1

Mass in grams = 0.1 mole x 36.5 g/mol

Mass in grams = 3.65 g

mass of HCl = 3.65 g

6 0

3 years ago

What is meant by the negative sign in an answer like"-46.8kJ"? When would you use a positive sign

MA_775_DIABLO [31]

It’s positive when you use energy for work

3 0

3 years ago

___CH4 + ___O2 → ___CO2 + ___H2O

DaniilM [7]

CH4 + 2O2 =======> CO2 + 2H2O

5 0

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