All categories

3 years ago

How the inside of a greenhouse works is an example of conduction, convection, or radiation?

Chemistry

2 answers:

faust18 [17]3 years ago

7 0

Radiation, the sun is radiating the inside of the greenhouse

gizmo_the_mogwai [7]3 years ago

6 0

Answer:

Radiation

Explanation: Because greenhouses have glass and sun travels through glass .

You might be interested in

If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

s2008m [1.1K]

The question is incomplete, the complete question is;

AgNO3(aq)+NaCl(aq)→AgCl(s)⏐↓+NaNO3(aq)

1- If 4.52g of the precipitate is formed, how many grams of sodium chloride reacted?

Answer:

1.8 g

Explanation:

Given the equation;

AgNO3(aq)+NaCl(aq)→AgCl(s)↓+NaNO3(aq)

We can see that the reaction is 1:1 Hence, 1 mole of sodium chloride yielded 1 mole of the precipitate(AgCl).

If this is so,

Number of moles of precipitate formed = 4.52g/143.32 g/mol

Number of moles of precipitate formed = 0.0315 moles

Hence, 0.0315 moles of precipitate was formed by 0.0315 moles of NaCl

Therefore;

Mass of NaCl reacted = 0.0315 moles * 58.5 g/mol = 1.8 g

7 0

2 years ago

What has an imcomplete octet in its lewis structure?

Leya [2.2K]

A bond can be drawn as a line between two atoms, which indicates two atoms

4 0

3 years ago

Rust is formed by the interaction of iron and oxygen. When rust formed on an iron nail with a mass of 100 g, the mass of the rus

DanielleElmas [232]

Answer: The correct option is A.

Explanation: Moles of Iron nail can be calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ...(1)

Given mass of iron nail = 100 g

Molar mass of iron = 56 g/mol

$\text{Number of moles}=\frac{100g}{56g/mol}$

Number of moles of iron nail = 1.79 moles

The chemical equation for the formation of rust if given by:

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

By Stoichiometry of the reaction,

4 moles of iron reacts with 3 moles of oxygen gas to produce 2 moles of rust.

So, 1.79 moles of iron will react with = $\frac{3}{4}\times 1.79$ = 1.3425 moles of oxygen gas.

For oxygen gas,

Molar mass = 32 g/mol

Mass of oxygen can be calculated by using equation 1, we get

$1.3425mol=\frac{\text{Given mass}}{32g/mol}$

Mass of oxygen gas = 43 grams

Hence, the correct option is a.

6 0

2 years ago

How many molecules are found in 90 grams of Al(OH)3?

jarptica [38.1K]

Answer:

6.95 x 10²³ molecules/particles

Explanation:

First we need to find the total Empirical Mass. We can find this by adding each element's mass together.

Al = 26.982,

O = 15.999

H = 1.008

26.982 + 3(15.999) + 3(1.008) = 78.003.

Now we divide by the mass given (90 grams).

90/78.003 = 1.153801777.

We then take that number and multiply it by avogadro's number (6.022 x 10²³)

1.153801777 x avogadro's number = 6.95 x 10²³

6 0

2 years ago

How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC$

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

4 0

2 years ago