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IgorLugansk [536]

3 years ago

8

In preparation for a blizzard, a city dispatches a crew to spread salt on the street surface of the bridges around town. The sal

t is spread on the streets because it will likely change the freezing point of water from ____ to ____.

1 answer:

timofeeve [1]3 years ago

4 0

Solid to liquid because it wil defrost

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Write the cell notation for the voltaic cell that incorporates each of the following redox reactions:(a) Al(s) + Cr³⁺(aq) → Al³⁺

Sladkaya [172]

The cell notation for the given voltaic cell is as follows

Al(s) | Al³⁺(aq) || Cr³⁺ (aq) | Cr(s)

<h3>What is a voltaic cell?</h3>

The device in which electrons are transferred through an external source is known as a voltaic cell.

The electron flow of a Voltaic cell allows redox reactions to produce energy in the form of electricity.

It converts the energy of spontaneous redox reaction into electrical energy

An anode, a cathode, and a salt bridge are all parts of a voltaic cell's setup. The salt bridge serves to neutralize the system.

To write the cell notation, identify the oxidation and reduction reactions.

At the anode, oxidation takes place

$Al(s) \longrightarrow Al^{3+} (aq) + 3e^-$

At the cathode, reduction takes place

$Cr^{3+} (aq) + 3e^- \longrightarrow Cr(s)$

While writing cell notation, the anode is written on the left, and the cathode is written on the right

A single vertical line differentiates between the phases and a double vertical line represents the salt bridge.

Now, write the cell notation of the given Voltaic cell

Al(s) | Al³⁺(aq) || Cr³⁺ (aq) | Cr(s)

Learn more about Voltaic cells:

brainly.com/question/14312871

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7 0

1 year ago

There are two beakers that contain the same liquid substance at the same temperature. The larger beaker is 1,000ml and the small

azamat

Answer:

Answer choice B

Explanation:

Since you do not know the volume of the liquid in each beaker, the one in the smaller beaker could have more substance and therefore more thermal energy. If they had the same amount of substance, then the more voluminous one would radiate faster. However, since you do not know this, there is no way to tell. PM me if you have more questions. Hope this helps!

6 0

2 years ago

when methane reacts with oxygen, the products are carbon dioxide and water. How many grams of carbon dilxide are formed if 30 g

Wittaler [7]

Answer:

CH4+2O2→CO2+2H2O

Moles of oxygen=3232=1

Moles of carbon dioxide=4422=0.5

Moles of water=1818=1

1 mole of methane reacts with 2 moles of oxygen to give 1 mole of CO2 and 2 moles of water so 0.5 mole of methane will be required which is equal to 0.5×16=8g.

Explanation:

8g

4 0

2 years ago

Help please asap. chem sucks

luda_lava [24]

I'm assuming that C is carbon.

$4.590 \: mol \: c \times \frac{12.01 \:g \: c}{1 \: mol \: c}$

55.1259 g of C

6 0

2 years ago

A compound is 42.9% C, 2.4% H, 16.7% N, and 38.1% O, by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, lowers the

Romashka [77]

This is an incomplete question, here is a complete question.

A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?

Answer : The molecular of the compound is, $C_6H_4N_2O_4$

Explanation :

First we have to calculate the mass of benzene.

$\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}$

$\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g$

Now we have to calculate the molar mass of unknown compound.

Given:

Mass of unknown compound (solute) = 6.45 g

Mass of benzene (solvent) = 43.95 g = 0.04395 kg

Formula used :

$\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}$

where,

$\Delta T_f$ = change in freezing point = $5.53-1.37=4.16^oC$

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of benzene

Molal-freezing-point-depression constant $(K_f)$ for benzene = $5.12^oC/m$

m = molality

Now put all the given values in this formula, we get

$4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}$

$\text{Molar mass of unknown compound}=180.6g/mol$

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 42.9 g

Mass of H = 2.4 g

Mass of N = 16.7 g

Mass of O = 38.1 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of N = 14 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles$

Moles of N = $\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.575}{1.193}=2.99\approx 3$

For H = $\frac{2.4}{1.193}=2.01\approx 2$

For N = $\frac{1.193}{1.193}=1$

For O = $\frac{2.381}{1.193}=1.99\approx 2$

The ratio of C : H : N : O = 3 : 2 : 1 : 2

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_3H_2N_1O_2$

The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{180.6}{84}=2$

Molecular formula = $(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4$

Therefore, the molecular of the compound is, $C_6H_4N_2O_4$

3 0

3 years ago