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2 years ago

How many molecules of calcium chloride are in 3CaCl2

Chemistry

1 answer:

Soloha48 [4]2 years ago

8 0

3*40+3*2*35.5=333* Avogadro’s number =
1.98*10^25

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Symbol for protons, neutrons, electrons in chemistry.

CaHeK987 [17]

This is the symbols for protons, neutrons, and electrons

7 0

3 years ago

Read 2 more answers

8. A sample of chloroform is found to contain 24.0 g of carbon,212.8 g of chlorine, and 2.02 g of hydrogen. If a second sample o

Misha Larkins [42]

Answer:

595.5

Explanation:

chloroform with 24.0 g C was 238.2 g

24g/238.2g= 60g/x

595.5

4 0

3 years ago

Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)

Gelneren [198K]

Answer:

ΔS = +541.3Jmol⁻¹K⁻¹

Explanation:

Given parameters:

Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹

Standard Entropy of C = 5.7Jmol⁻¹K⁻¹

Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹

Standard Entropy of CO = 198Jmol⁻¹K⁻¹

To find the entropy change of the reaction, we first write a balanced reaction equation:

Fe₂O₃ + 3C → 2Fe + 3CO

To calculate the entropy change of the reaction we simply use the equation below:

ΔS = ∑S $_{products}$ - ∑S $_{reactants}$

Therefore:

ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1

ΔS = +541.3Jmol⁻¹K⁻¹

5 0

3 years ago

How can Newton's Third Law of Motion be used to explain why it is difficult to walk on a slippery surface, such as ice?

Olenka [21]

E ground pushes you forward. But that interaction is friction. Reduce friction and it doesn't matter how strong your legs are, the surface is incapable of pushing you accordingly. The coefficient of static friction is very low so it is easy to slide your foot rather than push.

6 0

3 years ago

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did

user100 [1]

Answer:

4.5 L water we have in litres (L).

Explanation:

$Q=m\times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

Plugging in the values

$\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$$

So,

Volume of water = mass/density

$\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$$

=4.5 L (Answer)

6 0

3 years ago