All categories

2 years ago

Which milligram quantity contains a total of four significant figures

Chemistry

2 answers:

denis23 [38]2 years ago

7 0

D has a total of four significant figures.

Ksenya-84 [330]2 years ago

3 0

Answer:

Explanation:

The first 0 is sandwiched between the 3&1 and a ending 0 that is after the decimal is significant.

You might be interested in

Determine the theoretical yield of H2S (in moles) if 4.0 molAl2S3 and 4.0 mol H2O are reacted according to the followingbalanced

mamaluj [8]

Answer:

The theoretical yield of the hydrogen sulfide is 68.0 grams.

Explanation:

$Al_2S_3(s)+6H_2O(l)\rightarrow 2Al(OH)_3(s)+3H_2S(g)$

Moles of aluminum sulfide = 4.0 mol

Moles of water = 4.0 mol

According to reaction, 6 moles of water reacts with 1 mole of aluminum sulfide,then 4 moles of water will react with :

$\frac{1}{6}\times 4 mol=1.5 mol$ of aluminum sulfide

1.5 moles aluminum sulfide < 4 moles aluminum sulfide

This means that water is present in limiting amount and aluminum sulfide is in excess amount.So, amount of hydrogen sulfide will depend upon moles of water.

According to reaction, 6 moles of water gives with 3 mole of hydrogen sulfide,then 4 moles of water will give :

$\frac{3}{6}\times 4 mol=2.0 mol$ of hydrogen sulfide

Mass of hydrogen sulfide:

2.0 mol × 34 g/mol = 68.0 g

The theoretical yield of the hydrogen sulfide is 68.0 grams.

7 0

2 years ago

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this p

xxTIMURxx [149]

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

$Moles=1.20 \times {25.0\times 10^{-3}}\ moles$

Moles of potassium chloride = 0.03 moles

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=0.900 \times {15.0\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.0135 moles

According to the given reaction:

$2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

Theoretical yield = 3.75435 g

Given experimental yield = 2.45 g

% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %

6 0

2 years ago

Combustion of 8.652 g of a compound containing c, h, o, and n yields 11.088 g of coz, 3.780 g of h2o, and 3.864 g of no2. How ma

Katyanochek1 [597]

Answer:- C = 3.024 g, H = 0.42 g, N = 1.176 g and O = 4.032 g

Solution:- The compound contains C, H, N and O. On combustion, all the carbon is converted to carbon dioxide, All hydrogen is converted to water and all the nitrogen is converted to nitrogen dioxide.

From the grams of all these, we could calculate their moles and then using mol ratio of these products and the number of moles of C, H and N present in them, we calculate the moles of C, H and N respectively. Further, these moles are converted to grams. On subtracting the sum of grams of C, H and N from the mass of the sample, the mass of oxygen is calculated.

The calculations are as follows:

Calculations for the grams of C:-

$11.088gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})(\frac{12gC}{1molC})$

= 3.024 g C

Calculations for the grams of H:-

$3.780gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})(\frac{1gH}{1molH})$

= 0.42 g H

Calculations for the grams of N:-

$3.864gNO_2(\frac{1molNO_2}{46gNO_2})(\frac{1molN}{1molNO_2})(\frac{14gN}{1molN})$

= 1.176 g N

Mass of the compound is given as 8.652 g. Now we could calculate the grams of oxygen as:

mass of oxygen = 8.652 - (3.024 + 0.42 + 1.176)

= 8.652 - 4.62

= 4.032 g

So, 8.652 grams of the compound contains 3.024 g of C, 0.42 g of H, 1.176 g of N and 4.032 g of O.

7 0

2 years ago

Identify the oxidizing and reducing agents in the following: 2H+(aq) + H2O2(aq) + 2Fe2+(aq) → 2Fe3+(aq) + 2H2O(l)

schepotkina [342]

Answer : The oxidizing and reducing agents are, $H_2O_2$ and $Fe^{2+}$ .

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

The given redox reaction is:

$2H^+(aq)+H_2O_2(aq)+2Fe^{2+}(aq)\rightarrow 2Fe^{3+}(aq)+2H_2O(l)$

The half oxidation-reduction reactions are:

Oxidation reaction : $Fe^{2+}\rightarrow Fe^{3+}+1e^-$

Reduction reaction : $O^-+1e^-\rightarrow O^{2-}$

The oxidation state of oxygen in $H_2O_2$ and $H_2O$ is, (-1) and (-2) respectively.

In this reaction, 'Fe' is oxidized from oxidation (+2) to (+3) and 'O' is reduced from oxidation state (-1) to (-2). Hence, ' $Fe^{2+}$ ' act as a reducing agent and ' $H_2O_2$ ' act as a oxidizing agent.

Thus, the oxidizing and reducing agents are, $H_2O_2$ and $Fe^{2+}$ .

7 0

2 years ago

According to the law of conservation of mass in a chemical reaction the total starting mass of all the reactants equals the tota

Ber [7]

I believe the correct answer true. According to the law of conservation of mass, in a chemical reaction the total starting mass of all the reactants equals the total final mass of all the products. This law states that mass cannot be created or be destroyed. So, the total mass that goes in a process should be equal to the mass that goes out the process. This is true for chemical reactions and physical processes. It is Antoine Lavoisier who described this and is a basic principle used in physics and in chemistry. Mass, unlike energy, cannot be transformed to any form so however the transformation happens the mass should be constant.

3 0

2 years ago