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3 years ago

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Sharifa's car has broken down. She gets out and tries to push her car, but it will not move. Which type of friction is preventin

g her from moving her car?

1 answer:

Pachacha [2.7K]3 years ago

8 0

Inertia I believe but I’m not 100

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Compounds A and B react to form compounds C and D according to the equation: aA + bB → cC + dD. Under which conditions will the

Ipatiy [6.2K]

Answer: A. The reaction takes place in one step.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

$aA=bB\rightarrow cC+dD$

$Rate=k[A]^a[B]^b$

k= rate constant

a= order with respect to A

b = order with respect to B

5 0

3 years ago

Read 2 more answers

Definition: A compound composed of two elements that share valence electrons

Evgen [1.6K]

Answer:

Explanation:

Binary compound

3 0

3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

3 years ago

What is the molarity of a solution containing 60. grams of NaOH dissolved in 400. mL of water?

Hunter-Best [27]

Answer: 3.75 M

Explanation:

400 mL = 0.4 L

NaOH has a molar mass of around 40 g/mol.

$\frac{60 grams}{40 g/mol}$ = 1.5 moles

Molarity = $\frac{1.5 moles}{0.4 Liters}$ = 3.75 M

8 0

2 years ago

Re-order each list in the table below, if necessary, so that the atoms or ions in it are listed in order of decreasing size.

Serjik [45]

Answer: Na, S, Cl

Explanation:

Atomic size decreases as one moves from left to right on the periodic table with elements in the same period. This is as a result of the electrons increasing in the outer circle and thus being drawn to the protons in the nucleus which will lead to the outer shell area decreasing.

Sodium (Na) comes before Sulfur (S) which comes before Chlorine (S) so this is the decreasing order as they are all in the same period.

7 0

2 years ago