All categories

2 years ago

What is the volume of .750 M hydrochloride acid required to react completely with 25.00 mL of .290 M NaOH solution?

Chemistry

2 answers:

Whitepunk [10]2 years ago

8 0

Explanation:

lease please please please please please please

crimeas [40]2 years ago

7 0

from

HCl + NaOH----> NaCl +H2O

n of acid=1

n of base=1

Molarity of acid=750M

Molarity of base=290M

volume of base=25ml

from

Va=Mb ×vb×na

Ma×nb

Va=290×25×1

750×1

Va=7250

750

Va=9.67ml

You might be interested in

If u pour flour and sugar in one bowl is the substance a mixture atom or element

777dan777 [17]

Im pretty sure it's element

3 0

2 years ago

The element Oxygen, O, is an example of which of the following?

Anon25 [30]

It is an example of a molecule

5 0

3 years ago

Read 2 more answers

Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit

riadik2000 [5.3K]

Answer:

A) SiO2 is the limiting reactant

B) Theoretical yield= 72333.3g

C) % yield =91.5%

Explanation:

SiO2(s) + 2C(s) --------------> Si(s) + 2CO(g)

n(SiO2)= 155000/60 = 2583.33 mols

n(C)= 79000/12= 3291.66 mols

a)SiO2 is the limiting reactant

According to the balanced reaction equation,

60g of SiO2 produced 28g of SiO2

155000g of SiO2 will produce 155000×28/60= 72333.3g

Therefore theoretical yield of Si= 72333.3g

% yield= 66200/72333.3×100/1 =91.5%

5 0

3 years ago

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

3 0

2 years ago

Read 2 more answers

Which electronic configuration matches the model?

antiseptic1488 [7]

Answer:

option 2 is correct answer. its nitrogen.

7 0

2 years ago