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3 years ago

What is the percent by mass of Ca in calcium chloride, CaCI2 (show your work)

Chemistry

1 answer:

adoni [48]3 years ago

5 0

Answer:

$\boxed{ \tt{ \frac{Ca}{CaCl_2} \times \frac{100}{1} }}$

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What is the change in internal energy ( ΔU ) of the system if q = –8 kJ and w = –1 kJ for a certain process?

Radda [10]

Answer:

Change in internal energy (ΔU) = -9 KJ

Explanation:

Given:

q = –8 kJ [Heat removed]

w = –1 kJ [Work done]

Find:

Change in internal energy (ΔU)

Computation:

Change in internal energy (ΔU) = q + w

Change in internal energy (ΔU) = -8 KJ + (-1 KJ)

Change in internal energy (ΔU) = -8 KJ - 1 KJ

Change in internal energy (ΔU) = -9 KJ

6 0

3 years ago

If you burn 50.6 g of hydrogen and produce 452 g of water, how much oxygen reacted?

Svetllana [295]

Acc. to Law of Conservation of Mass
Mass of reactants=Mass of Products
Let mass of Oxygen be x.
So,
50.6+x=452
x=452-50.6
=401.4 g

4 0

4 years ago

Which electron dot diagram represents H2?

Marizza181 [45]

Answer:

H:H

Explanation:

7 0

3 years ago

A 13 g sample of P4010 contains how many

Alex777 [14]

Answer:

$\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}$

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅, then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

$\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}$

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

$\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}$

3. Molecules of P₂O₅

$\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}$

6 0

4 years ago

Calculate the rate constant at 200.°C for a reaction that has a rate constant of 8.30 × 10−4 s−1 at 90.°C and an activation ener

Sergeu [11.5K]

Answer:

23.0 s⁻¹ is rate constant

Explanation:

Using the Arrhenius equation:

k = A * e^(-Ea/RT)

Where k is rate constant

A is frequency factor (1.5x10¹¹s⁻¹)

Ea is activation energy = 55800J/mol

R is gas constant (8.314J/molK)

And T is absolute temperature (24°C + 273 = 297K)

Replacing:

k = 1.5x10¹¹s⁻¹ * e^(-55800J/mol/8.314J/molK*297K)

k = 1.5x10¹¹s⁻¹ * 1.53x10⁻¹⁰

k = 23.0 s⁻¹ is rate constant i hope this helpsss

Explanation:

4 0

3 years ago