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3 years ago

Question 1(Multiple Choice Worth 4 points)

Chemistry

2 answers:

lys-0071 [83]3 years ago

4 0

Answer:

Question 3: 4/2

Explanation:

I just took the test. ;)

zubka84 [21]3 years ago

3 0

<u>Answer </u>

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

<u>Solution </u>

Solution 1 : Given,

Given mass of $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ gives → 3 × 0.344 moles of CO

= 1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$

The balanced chemical equation is,

$Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}$

From the given balanced equation, we conclude that

1 mole of $Zn(ClO_{3})_{2}$ gives → 3 moles of $O_{2}$

$(\frac{150\times 1}{232.29})moles$ of $Zn(ClO_{3})_{2}$ gives → $[(\frac{150\times 1}{232.29})\times 3] moles$ of $O_{2}$

Mass of $O_{2}$ = Number of moles of $O_{2}$ × Molar mass of $O_{2}$ = $[(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams$

Therefore, the mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of $Na_{2}SO_{4}$ = 4.2 moles

Balanced chemical equation is,

$H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}$

From the given chemical reaction, we conclude that

1 mole of $Na_{2}SO_{4}$ obtained from 2 moles of NaCN

4.2 moles of $Na_{2}SO_{4}$ obtained → 2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles

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combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon

masya89 [10]

Answer:

$\rm CH_2$ .

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be $\rm CO_2$ and $\rm H_2O$ . The $\rm C$ and $\rm H$ would come from the hydrocarbon, while the $\rm O$ atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

$\rm C$ : $12.011$ .
$\rm H$ : $1.008$ .
$\rm O$ : $\rm 15.999$ .

Calculate the molar mass of $\rm CO_2$ and $\rm H_2O$ :

$M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

$M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}$

Calculate the number of moles of $\rm CO_2$ molecules in $33.01\; \rm g$ of $\rm CO_2\!$ :

$\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol$ .

Similarly, calculate the number of moles of $\rm H_2O$ molecules in $13.51\; \rm g$ of $\rm H_2O\!$ :

$\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol$ .

Note that there is one carbon atom in every $\rm CO_2$ molecule. Approximately $0.7501\; \rm mol$ of $\rm CO_2\!$ molecules would correspond to the same number of $\rm C$ atoms. That is: $n(\mathrm{C}) \approx 0.7501\; \rm mol$ .

On the other hand, there are two hydrogen atoms in every $\rm H_2O$ molecule. approximately $0.7499\; \rm mol$ of $\rm H_2O$ molecules would correspond to twice as many $\rm H\!$ atoms. That is: $n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol$ .

The ratio between the two is: $n(\mathrm{C}): n(\mathrm{H}) \approx 1:2$ .

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be $\rm CH_2$ .

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3 years ago

What is defined as the basic unit in the metric system that describes a quality

Bingel [31]

Answer:

They are:

Length - meter (m)

Time - second (s)

Amount of substance - mole (mole)

Electric current - ampere (A)

Temperature - kelvin (K)

Luminous intensity - candela (cd)

Mass - kilogram (kg)

3 0

3 years ago

What volume of water is required to prepare 0.1 M H3PO4 from 100 ml of 0.5 M solution?

ExtremeBDS [4]

Answer: A volume of 500 mL water is required to prepare 0.1 M $H_{3}PO_{4}$ from 100 ml of 0.5 M solution.

Explanation:

Given: $M_{1}$ = 0.1 M, $V_{1}$ = ?

$M_{2}$ = 0.5 M, $V_{2}$ = 100 mL

Formula used to calculate the volume of water is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.1 M \times V_{1} = 0.5 M \times 100 mL\\V_{1} = 500 mL$

Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M $H_{3}PO_{4}$ from 100 ml of 0.5 M solution.

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3 years ago

To test paint quality, scientists designed an experiment where the paint was exposed to temperatures ranging from 260 degrees Ce

pishuonlain [190]

Answer:

The independent variable is the temperature to which the paint and materials are exposed

The dependent variable is the presence of peeling and cracking of the paint

The constant rates are the paint and material used for each trial

Explanation:

The treatment the to which the paint is subjected = Temperatures from 260 °C to 650 °C

The method of determining the paint quality = By peeling and cracking

The types of paint and materials used = The same paint and material

The independent variable is the variable suspected as the cause of the effect being studied. It is the variable introduced as a treatment

In the question, given that the paint is subjected to different temperatures from between 260°C to 650 °C, the temperature changes introduced is the independent variable

Therefore;

The independent variable = The temperature to which the paint and materials are exposed to

The dependent variable is the effect variable. It is the variable of the property being investigated and it is also the variable that is measured in the investigation

In the question, given that the after heating the paint and material, the scientist check for peeling and cracking, the dependent variable is the occurrence of cracking and peeling at a given temperature

Therefore;

The dependent variable = The presence of peeling and cracking of the paint

The constant rate(s) are variables which are kept constant during the investigation, under the different treatments

The constant rate(s) in the tests are the paint and material, which were kept the same for each trial to reduce the effect of underlying factors that may impact on the result and obscure the relationship between the dependent and independent variables

Therefore;

The constant rate are the paint and material used for each trial

8 0

3 years ago

Consider a cobalt-silver voltaic cell that is constructed such that one half-cell consists of the cobalt, Co, electrode immersed

umka21 [38]

Answer:

Anode half reaction;

Co(s) ----> Co^2+(aq) + 2e

Cathode half reaction;

2Ag^+(aq) + 2e-------> 2Ag(s)

Explanation:

A voltaic cell is an electrochemical cell that spontaneously produces electrical energy from chemical reactions. A voltaic cell comprises of an anode (where oxidation occurs) and a cathode (where reduction occurs). The both electrodes are connected with a wire . A salt bridge ensures charge neutrality in the anode and cathode compartments. Electrons flow from anode to cathode.

For the cell referred to in the question;

Anode half reaction;

Co(s) ----> Co^2+(aq) + 2e

Cathode half reaction;

2Ag^+(aq) + 2e-------> 2Ag(s)

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3 years ago