A bar of soap is 12 cm long, 6 cm wide, and 10 cm long. It has a mass of 415 grams. What is the density of the bar of soap?

Helium gas in a cylinder is under 1.12atm pressure at 25.0C. What will be the pressure if the temperature increases to 37.0C?

Irina18 [472]

Answer:

$p_2=1.17atm$

Explanation:

Hello!

In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:

$\frac{p_2}{T_2} =\frac{p_1}{T_1} \\\\$

Thus, for the initial pressure and temperature in kelvins the final temperature in kelvins, we compute the final pressure as:

$p_2=\frac{p_1T_2}{T_1} \\\\p_2=\frac{1.12atm*310.15K}{298.15K}\\\\p_2=1.17atm$

Best regards!

7 0

2 years ago

How many moles of Li would react with 4 moles of MgBr₂ in the reaction

ad-work [718]

Answer:

two may be the answer (2)

3 0

2 years ago

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with

emmasim [6.3K]

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

It is a stichiometry problem.

2Na + 2H₂O → 2NaOH + H₂,

The balanced equation shows that 2.0 moles of Na metal react with 2.0 moles of water to produce 2.0 moles of NaOH and 1.0 mole of H₂,

Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: PV = nRT,

where, P is the pressure of the gas in atm (P at STP = 1.0 atm),

V is the volume of the gas in L (V = 7.80 L),

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K (T at STP = 0.0 °C + 273 = 273.0 K).

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

Using cross multiplication:

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.

6 0

3 years ago

All help is needed, thanks

yarga [219]

Answer:

smoke
marshmallow
paint
Body spray

Explaination:

Firstly, what are colloids?

Colloids are a mixture of two substances:

one of those substances are insoluble and inseparable

There are 4 types of colloids:

Sol- Solid in a liquid
Emulsion- Liquid in a liquid
Foam- Gas in solid/liquid
Aerosol- Solid/liquid in gas

So on to the answers:

Smoke consists of solid and gas and since the solid is being carried by the gas, it is a solid in a gas (aerosol)
Marshmellow consists of tiny air bubbles and sugar(solid), the air bubbles is inside the solid making it a gas in a solid (foam)
Body spray turns gas into liquid after spraying it through the can, making it a liquid in a gas (aerosol)
Paint dries up to become solid which makes it a solid is a liquid (sol)

Hope y'all understand

3 0

3 years ago