If Justin races his Chevy camaro south down 1-540 for 2560 meter in 60 second.what is it velocity

7. A figure skater skating across ice, who grabs another skater and brings him along the ice with her, is an example of what typ

xenn [34]

Answer:

The type of collision is A. Inelastic collision.

Explanation:

The macroscopic collisions are generally inelastic and do not conserve the kinetic energy, although of course the total energy is conserved. The inelastic collision is one in which the objects that collide remain together after the collision.

So, a figure skater skating across ice, who grabs another skater and brings him along the ice with her is a clear example of inelastic collision.

6 0

3 years ago

Read 2 more answers

What volume of carbon dioxide gas can be collected from

alisha [4.7K]

Answer:

1.22 L of carbon dioxide gas

Explanation:

The reaction that takes place is:

CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O

First we <u>determine which reactant is limiting</u>:

Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃

Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl

So HCl is the limiting reactant.

Now we calculate the moles of CO₂ produced:

0.05 mol HCl * $\frac{1molCO_{2}}{1molHCl}$ = 0.05 mol CO₂

Finally we use PV=nRT to <u>calculate the volume</u>:

P = 1 atm

n = 0.05 mol

T = 25 °C ⇒ 25 + 273.16 = 298.16 K

1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

V = 1.22 L

7 0

3 years ago

5. What concentration of acid must be added to change the pH of 1 mM phosphate buffer from 7.4 to 7.3 (pKas of the phosphate buf

mr_godi [17]

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log \frac{base}{acid}$

where, pH = 7.4 and $pK_{a}$ = 7.21

As here, we can use the $pK_{a}$ nearest to the desired pH.

So, 7.4 = 7.21 + $log \frac{base}{acid}$

0.19 = $log \frac{base}{acid}$

$\frac{base}{acid}$ = 1.55

1 mM phosphate buffer means $[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM

Therefore, the two equations will be as follows.

$\frac{HPO_{4}}{H_{2}PO_{4}}$ = 1.55 ............. (1)

$[HPO_{4}]$ + $[H_{2}PO_{4}]$ = 1 mM ........... (2)

Now, putting the value of $[HPO_{4}]$ from equation (1) into equation (2) as follows.

1.55 $[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]$ = 1 mM

2.55 $[H_{2}PO_{4}]$ = 1 mM

$[H_{2}PO_{4}]$ = 0.392 mM

Putting the value of $[H_{2}PO_{4}]$ in equation (1) we get the following.

0.392 mM + $[HPO_{4}]$ = 1 mM

$[HPO_{4}]$ = (1 - 0.392) mM

$[HPO_{4}]$ = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

7 0

2 years ago

There are millions of organic compounds but only thousands of inorganic compounds because:______

siniylev [52]

Answer:

There are millions of organic compounds but only thousands of inorganic compounds because:

a. organic compounds were formed by living things.

b. there is more carbon on Earth's surface than any other element.

c. atoms of elements other than carbon never combine with themselves.

d. carbon atoms can combine with up to four other atoms, including other carbon atoms.

6 0

1 year ago

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant

antoniya [11.8K]

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

$K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

$Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$ , $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$ , $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

= $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

$E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

$0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$

$E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

= $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

= 0.577 V

Hence, we will calculate the standard cell potential as follows.

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

$0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

$0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

$E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.

3 0

2 years ago