____________ enables you to ride a bike without skidding and falling.

Help me I need to know how do do this or just give me answers lol

Pani-rosa [81]

4. 2Li + 2H2O -> 2LiOH + H2
5. C6H12O6 + 6O2 -> 6CO2 + 6H2O
6. Zn + 2HCl -> ZnCl2 + H2
9. H2SO4 + Pb -> PbSO4 + H2
10. Ca(OH)2 + NH4Cl -> NH4 + CaCl2 + H2O
thats all i know

5 0

3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio

evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

$CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)$

We can write the law of mass action for it:

$Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}$

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change $x$ due to the dissociation extent, we are able to rewrite it as shown below: