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4 years ago

If I raise the Celsius scale by 15 degrees, how much will I be raising it in the Kelvin Scale?

Chemistry

1 answer:

Mariana [72]4 years ago

6 0

Answer:

If the change is +15 degrees Celsius, the change is +15 Kelvin (note that it's not degrees Kelvin, just simply Kelvin). You can use the conversion, which is 273.15 Kelvin = zero degrees Celsius. Since it's a relative scale, 10 C = 274.15 K, and 20 C = 275.15 and so forth.

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Need help asap!!!!! Plz

Nostrana [21]

Answer:

153.8 g/mol.

Explanation:

The molar mass of a compound = ∑ atomic masses of atoms.

∴ The molar mass of CCl₄ = 1(atomic masss of C) + 4 (atomic mass of Cl) = (12.0 g/mol) + 4 (35.45 g/mol) = 153.8 g/mol.

6 0

3 years ago

Calculate the mass of arsenic trihydride produced from 117.4 g of hydrogen gas with excess arsenic trioxide:

Lana71 [14]

Answer: The mass of $AsH_3$ produced is, 1528.8 grams.

Explanation : Given,

Mass of $H_2$ = 117.4 g

Molar mass of $H_2$ = 2 g/mol

First we have to calculate the moles of $H_2$ .

$\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}$

$\text{Moles of }H_2=\frac{117.4g}{2g/mol}=58.7mol$

Now we have to calculate the moles of $AsH_3$

The balanced chemical equation is:

$6H_2+As_2O_3\rightarrow 2AsH_3+3H_2O$

From the reaction, we conclude that

As, 6 moles of $H_2$ react to give 2 moles of $AsH_3$

So, 58.7 moles of $H_2$ react to give $\frac{2}{6}\times 58.7=19.6$ mole of $AsH_3$

Now we have to calculate the mass of $AsH_3$

$\text{ Mass of }AsH_3=\text{ Moles of }AsH_3\times \text{ Molar mass of }AsH_3$

Molar mass of $AsH_3$ = 78 g/mole

$\text{ Mass of }AsH_3=(19.6moles)\times (78g/mole)=1528.8g$

Therefore, the mass of $AsH_3$ produced is, 1528.8 grams.

6 0

3 years ago

How do you calculate wave speed?

Goshia [24]

For a wave, speed= frequency x wavelength

8 0

3 years ago

The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g)

wolverine [178]

Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478. The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59

A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.

6 0

3 years ago

The point at where the water is changing phase

e-lub [12.9K]

Answer:

the point at there the water is changing is titled "getting dressed"

6 0

3 years ago