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Musya8 [376]
2 years ago
15

How many grams of mgcl2 are contained in 500 grams of a 1.0% solution

Chemistry
1 answer:
GarryVolchara [31]2 years ago
5 0

1. Julie had saved 105 coins from her daily allowance. If the coins consist of 5-peso coins and 10-peso coins amounting to ₱950, how many of each kind of coin did she have?

1. Julie had saved 105 coins from her daily allowance. If the coins consist of 5-peso coins and 10-peso coins amounting to ₱950, how many of each kind of coin did she have?

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
How many dm³ of hydrogen,measured at s.t.p.,would be needed to reduce 47.7g of copper(II) oxide to copper?
GaryK [48]

Answer:

Option D. 13.44

Explanation:

We'll begin by calculating the number of mole in 47.7g of copper(II) oxide, CuO.

This can be obtained as follow:

Mass of CuO = 47.7 g

Molar mass of CuO = 63.5 + 16 = 79.5 g/mol

Mole of CuO =.?

Mole = mass /Molar mass

Mole of CuO = 47.7/79.5

Mole of CuO = 0.6 mole

Next, we shall write the balanced equation for the reaction. This is given below:

CuO + H2 —> Cu + H2O

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2 to produce 1 mole of Cu and 1 mole of H2O.

Next, we shall determine the number of mole of H2 needed to react completely with 0.6 mole of CuO.

This can be obtained as follow:

From the balanced equation above,

1 mole of CuO reacted with 1 mole of H2.

Therefore, 0.6 mole of CuO will also react with 0.6 mole of H2.

Finally, we shall determine the volume occupied by 0.6 mole of H2 at STP.

This can be obtained as follow:

1 mole of H2 occupied 22.4 dm³ at STP.

Therefore, 0.6 mole of H2 will occupy = 0.6 × 22.4 = 13.44 dm³.

Therefore, 13.44 dm³ of H2 is needed for the reaction.

4 0
2 years ago
Do subatomic particles have mass
katovenus [111]

Answer:

Yes

Explanation:

Masses for the three subatomic particles can be expressed in amu (atomic mass units) or grams. For simplicity, we will use the amu unit for the three subatomics. Both neutrons and protons are assigned as having masses of 1 amu each.

5 0
3 years ago
Which one is you fav one
serious [3.7K]

Answer:

the 3rd one

Explanation:

8 0
2 years ago
Read 2 more answers
How can I balance this equation? ____ KClO3 ---> ____ KCl + ____ O2
tensa zangetsu [6.8K]
__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂
8 0
3 years ago
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