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Oxidation state of Nitrogen in N2O5

balandron [24]

Explanation:

Oxidation state of Nitrogen in N2O5 is +5

7 0

2 years ago

If a chemical has a pH of 3, how could you alter its pH value to be more basic?

Andreyy89

Answer; If a chemical has a pH of 3, how could you change its pH value to be more basic? Adding water to a chemical will dilute the acid, thus lowering the pH value to more basic.

6 0

3 years ago

in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?

Romashka-Z-Leto [24]

Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

4 0

3 years ago

Assume the molality of isoborneol in your product is 0.275 mol/kg. What is the melting point of your impure sample given that th

aliina [53]

Answer:

168°C is the melting point of your impure sample.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = ?

Depression in freezing point = $\Delta T_f$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample = 0.275 mol/kg

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( non electrolyte)

$\Delta T_f=1\times 40^oC kg/mol\times 0.275 mol/kg$

$\Delta T_f=11^oC$

$\Delta T_f=T- T_f$

$T_f=T- \Delta T_f=179^oC-11^oC=168^oC$

168°C is the melting point of your impure sample.

4 0

3 years ago

In going from room temperature (25 C) to 10 C above room temperature, the rate of reaction doubles. Calculate the activation ene

Sauron [17]

Answer:

Ea=5.29 × 10⁴ J/mol

Explanation:

In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol$

8 0

3 years ago