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2 years ago

Which of the following statements about tornadoes is false?

Chemistry

2 answers:

loris [4]2 years ago

4 0

The answer is A because the time wouldn't affect the weather that much depending on the circumstances.

Rainbow [258]2 years ago

4 0

Answer:

Explanation:

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The chemical equation below shows the formation of aluminum oxide (Al2O3) from aluminum (Al) and oxygen (O2).

Rufina [12.5K]

Answer:

Mass = 182.4 g

Explanation:

Given data:

Number of moles of Al₂O₃ = 3.80 mol

Mass of oxygen required = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Now we will compare the moles of aluminum oxide and oxygen.

Al₂O₃ : O₂

2 : 3

3.80 : 3/2×3.80 = 5.7

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 5.7 mol × 32 g/mol

Mass = 182.4 g

4 0

2 years ago

Which mass of urea, CO(NH2)2, contains the same mass of nitrogen as 101.1g of potassium nitrate?

Effectus [21]

In order to calculate the mass of nitrogen, we must first calculate the mass percentage of nitrogen in potassium nitrate. This is:
% nitrogen = mass of nitrogen / mass of potassium nitrate
% nitrogen = 14 / 101.1 x 100

The mass of nitrogen = % nitrogen x sample mass
= (14 / 101.1) x 101.1
= 14 grams

The molar weight of nitrogen is 14. Each mole of urea contains two moles of nitrogen. Therefore, for there to be 14 grams of nitrogen, there must be 0.5 moles of urea.
Mass of urea = moles urea x molecular weight urea
Mass of urea = 0.5 x 66.06
Mass of urea = 33.03 grams

4 0

2 years ago

The energy of a pendulum is recorded at different positions in the table below. Use the data in the table to determine the amoun

natta225 [31]

Answer:

36.55 J

Explanation:

PE = Potential energy

KE = Kinetic energy

TE = Total energy

The following data were obtained from the question:

Position >> PE >>>>> KE >>>>>> TE

1 >>>>>>>> 72.26 >> 27.74 >>>> 100

2 >>>>>>>> 63.45 >> x >>>>>>>> 100

3 >>>>>>>> 58.09 >> 41.91 >>>>> 100

The kinetic energy of the pendulum at position 2 can be obtained as follow:

From the table above, at position 2,

Potential energy (PE) = 63.45 J

Kinetic energy (KE) = unknown = x

Total energy (TE) = 100 J

TE = PE + KE

100 = 63.45 + x

Collect like terms

100 – 63.45 = x

x = 36.55 J

Thus, the kinetic energy of the pendulum at position 2 is 36.55 J.

7 0

2 years ago

Which element forms an ionic bond when combined with Cl

Shtirlitz [24]

Potassium or any other metals.

3 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

3 years ago