All categories

3 years ago

Choose all the answers that apply.

Chemistry

2 answers:

SVETLANKA909090 [29]3 years ago

8 0

Answer:

A. burning of fossil fuels

B. respiration

Explanation:

Fossil fuels such as petrol, diesel and coal are used in the movement of cars. These fuels are burnt inside the engine of vehicles and carbondioxide is released. Respiration is energy releasing process in which energy is released in the form of Adenine tri phosphate. In respiration, carbondioxide is released when glucose is broken down inside the mitochondria of the cell with the addition of oxygen.

LenKa [72]3 years ago

3 0

Burning of fossil fuels: when we burn fossil fuels, CO2 is produced

Respiration: when animals respire, they convert O2 to CO2 which is then released into the atmopshere

Hope it helps :)

You might be interested in

Liquids and gases are matter because they take up space and have

Aneli [31]

Volume. Gases and liquids are typically measured in milliliters (mL) or cubic centimeters (cm^3) - both of which are equivalent (1 mL = 1 cm^3).

8 0

3 years ago

Many types of scientific equipment are used to perform different functions in the science lab. Which of the following combinatio

Phantasy [73]

I guess B is the right one......

4 0

3 years ago

Read 2 more answers

How many moles are in 1.5 x 10^23 atoms of fluorine?

andrezito [222]

Answer:

0.052636002587839 moles

Explanation:

im not sure so sorry if u get it wrong...

4 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago

What element is in group 17 and period 3 of the periodic table?

Anastasy [175]

Answer:

Chlorine is the element located in period 3, group 17 of the periodic table.

Explanation:

6 0

3 years ago

Read 2 more answers