Samantha is using a lever to lift a rock. To life the rock most easily, where should she apply force to the lever

Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO

Mrac [35]

I'm not 100% sure on this, but I would go with C) NaCl.
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!

7 0

2 years ago

A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

2 years ago

mastering chem If the experiment below is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? At

Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

$\delta moles/\delta time = 0.30-0.16/60-40 = 0.007 Mol/s$ which is true as well

7 0

2 years ago

The reaction between hydrogen gas and chlorine gas produces hydrogen chloride according to the following equation: H2(g) + Cl2(g

Andreyy89

Answer:

The enthalpy of reaction is -185 kJ

Explanation:

To get the reaction:

H₂(g) + Cl₂(g) → 2 HCl(g)

you must follow the following steps:

1) Reactive molecules must break their bonds to obtain their atoms.

H₂(g) → 2 H(g)

Cl₂(g) → 2 Cl(g)

Bond energy (or enthalpy) is the energy required to break one mole of bonds of a gaseous substance. In the case of diatomic molecules with a single bond, it corresponds to the energy necessary to dissociate 1 mole of said substance in the atoms that form it.

Whenever you want to break links you must supply energy, so the link enthalpy will have positive values; while when a mole of bonds is formed energy is released and the bond enthalpy of this process will be negative.

In this case you will then have:

H₂(g) → 2 H(g) ΔH=436 kJ/mol

Cl₂(g) → 2 Cl(g) ΔH=243 kJ/mol

So the total energy needed to break all the bonds is:

ΔH=1 mol*436 kJ/mol +1 mol* 243 kJ/mol= 679 kJ

2) The atoms that were obtained in the break of the bonds must be combined to obtain the product.

2 H (g) + 2 Cl (g) → 2 HCl (g)

Being the single bond energy for one mole of 431 kJ H-Cl bonds and considering that two moles of H-Cl bonds are formed, the ΔH is:

ΔH = -2 moles* (432 kJ/mol) = -864 kJ

As mentioned, when a mole of bonds is formed energy is released, the bond enthalpy of this process will be negative. So the formation of HCl is negative.

Hess's law states that the energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it. So:

ΔHtotal= -864 kJ + 679 kJ

ΔHtotal= -185 kJ

The enthalpy of reaction is -185 kJ

3 0

2 years ago

Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80

AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0

2 years ago