What did ancient astronomers think areas of the moon called mares might be?

A scientist analyzes four samples of compounds. He finds the first compound contains 10 grams of sodium, 211 grams of potassium,

Zarrin [17]

Answer:

yea yea

Explanation:

6 0

3 years ago

Rank the following substances/solutions in order of lowest boiling point to highest boiling point where 1 has the lowest boiling

poizon [28]

Answer:

1) pure water

2) 0.75 m CaCl2

3) 1.0 m NaCl

4) 0.5 m KBr

5) 1.5 m glucose (C6H12O6)

Explanation:

Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.

The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.

The order of decreasing hoping point elevation is;

1) 0.75 m CaCl2

2) 1.0 m NaCl

3) 0.5 m KBr

4) 1.5 m glucose (C6H12O6)

5 0

3 years ago

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

8 0

3 years ago

Burning soybean oil is similar to what other chemical reaction?

Harrizon [31]

Answer:

Hydrogenation of the vegetables Oils .

Explanation:

8 0

3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

belka [17]

10 cubic inches
We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.
P1 V1 = P2 V2
Where
P1 is initial pressure = 5 psi
V1 is initial volume = 20 cubic inch
P2 is final pressure = 10 psi
V2 is final volume = unknown
V2 = P1,V1 / P2
V2 = 20 × 5 / 10
V2 = 100/10
V2 = 10 cubic inches

7 0

3 years ago