What is the mole of 98 mL of carbon dioxide gas at 36°C and 795 torr?

In the winter, a heated home in the Northeast might be maintained at a temperature of 78°F. What is this temperature on the Cels

masya89 [10]

Answer: $25.6^0C$ , $298.6K$

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like $^0C$ and $K$

These units of temperature are inter convertible.

We are given:

Temperature of the gas = $78^0F$

Converting this unit of temperature into $^0C$ by using conversion factor:

$^oC=\frac{5}{9}\times (^oF-32)$

$^0C=\frac{5}{9}\times (78^oF-32)$

$25.6^0C$

Converting this unit of temperature into $K$ by using conversion factor:

$K=t^0C+273$

$K=25.6+273$

$298.6K$

Thus the temperature on the Celsius and Kelvin scales are $25.6^0C$ and $298.6K$ respectively.

6 0

3 years ago

The km of an enzyme is 5. 0 mm. Calculate the substrate concentration when this enzyme operates at one‑quarter of its maximum ra

Makovka662 [10]

The substrate concentration of the enzyme operating at one‑quarter of its maximum rate is = 0.333.

Relationship between Km and substrate concentration is -

Km is the concentration of substrate.It allows the enzyme to achieve half Vmax. High Km enzyme requires a higher concentration of substrate to get Vmax. Since, Km is a constant. If the substrate concentration is increased, it has no effect on it.

An enzyme with a high Km has a low affinity for its substrate. The substrate concentration Km corresponds to the substrate concentration.

The substrate concentration at which the reaction rate of the enzyme-catalyzed reaction is half of the maximum reaction rate Vmax.

The equation is:

V₀ = Vmax [S]

[S] + Km

Here,

V₀ is initial rate,

Km is the dissociation constant between the substrate and the enzyme,

Vmax is the maximum rate, and

S is the concentration of substrate.

taking fraction of V₀ and Vmax :

V₀ = [S]

Vmax [S] + Km

V₀ = 0.5Km = 0.333

Vmax 1.50 + Km

Therefore, the substrate concentration of this enzyme operating at one‑quarter of its maximum rate is = 0.333.

To learn more about substrate concentration,

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3 0

10 months ago

Calculate the number of oxygen atoms and its mass in 50g of CaCo3

xz_007 [3.2K]

The number of oxygen atoms = 3

Mass = 24 g

<h3>Further explanation</h3>

The formula of a compound shows the composition of the constituent elements

CaCO₃ is composed of 3 types of elements, namely Ca, C and O

The amounts of each of these elements in the compound CaCO₃:

Ca = 1
C = 1
O = 3

So the number of oxygen atoms = 3

mass of Oxygen :

$\tt mass~O=\dfrac{3\times Ar~O}{MW~CaCO_3}\times mass~CaCO_3\\\\mass~O=\dfrac{3\times 16}{100}\times 50=24~g$

5 0

2 years ago

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

4 0

3 years ago

How do I do scientific notation

lina2011 [118]

1. First, move the decimal place until you have a number between 1 and 10. If you keep moving the decimal point to the right in 0.0000073 you will get 7.3.
2. Next, count how many places you moved the decimal point.

7 0

3 years ago