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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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How many milliliters of 0.1 m hcl is required to react with 0.15g sodium carbonate (na2co3?

docker41 [41]

1) Write the balanced chemical equation

2HCl + Na2 CO3 ----------> 2NaCl + H2CO3

2) Write the molar ratios:

2 mol HCl : 1 mol Na2CO3 : 2 mol NaCl : 1 mol H2CO3

3) Convert 0.15g of sodium carbonate to number of moles

3a) Calculate the molar mass of Na2CO3

Na: 2 * 23 g/mol = 46 g/mol

C: 12 g/mol =

O: 3 * 16 g/mol = 48 g/mol

molar mass = 46g/mol + 12g/mol + 48g/mol = 106 g/mol

3b.- Calculate the number of moles of Na2CO3

# moles = grams / molar mass = 0.15 g / 106 g/mol = 0.0014 mol Na2CO3

4) Calculate the number of moles of HCl from the molar proportion:

[0.0014 mol Na2CO3] * [2 mol HCl / 1 mol Na2CO3] = 0.0028 mol HCl

5) Calculate the volume of HCl from the definition of Molarity

Molarity, M = # moles / volume in liters

=> Volume in liters = # moles / M = 0.0028 mol / 0.1 M = 0.028 liters

0.028 liters * 1000 ml / liter = 28 ml.

Answer: 28 mililiters of 0.1 M HCl.

7 0

3 years ago

How many moles are present in

olasank [31]

Answer: (a) There are 0.428 moles present in 12 g of $N_{2}$ molecule.

(b) There are 2 moles present in $12.044 \times 10^{23}$ particles of oxygen.

Explanation:

(a). The mass of nitrogen molecule is given as 12 g.

As the molar mass of $N_{2}$ is 28 g/mol so its number of moles are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{12 g}{28 g/mol}\\= 0.428 mol$

So, there are 0.428 moles present in 12 g of $N_{2}$ molecule.

(b). According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

Therefore, moles present in $12.044 \times 10^{23}$ particles are calculated as follows.

$Moles = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}}\\= 2 mol$

So, there are 2 moles present in $12.044 \times 10^{23}$ particles of oxygen.

4 0

2 years ago

How is an oxidation half-reaction written using the reduction potential chart? How is the oxidation potential voltage determined

insens350 [35]

Oxidation half reaction is written as follows when using using reduction potential chart
example when using copper it is written as follows
CU2+ +2e- --> c(s) +0.34v
oxidasation is the loos of electron hence copper oxidation potential is as follows
cu (s) --> CU2+ +2e -0.34v

7 0

3 years ago

Read 2 more answers

Explain how nuclear and chemical reactions differ in terms of how they are balanced. (PLEASE ANSWER ASAP THIS NEEDS TO BE TURNED

Tcecarenko [31]

Explanation:

(1) Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei. ... (3) Rates of chemical reactions are influenced by temperature and catalysts.

6 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago