If two pretty beat friends exist would they actually be real or just 2 dogs 1.yes 2.no

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5g of na2s and

tangare [24]

The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.

8 0

4 years ago

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What is the molarity of a solution in which 7.1 g of sodium sulfate is dissolved in enough water to make 100. mL of solution?

Sidana [21]

Answer:

0.50 M

Explanation:

Given data

Mass of sodium sulfate (solute): 7.1 g

Volume of solution: 100 mL

Step 1: Calculate the moles of the solute

The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:

$7.1g \times \frac{1mol}{142.02g} = 0.050mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1 L = 1000 mL.

$100mL \times \frac{1L}{1000mL} =0.100L$

Step 3: Calculate the molarity of the solution

$M = \frac{moles\ of\ solute }{liters\ of\ solution} = \frac{0.050mol}{0.100L} =0.50 M$

3 0

3 years ago

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Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a

Annette [7]

<h3>solution:</h3>

The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:

s= 8.2652 kJ/kgK

The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:

T₂ = T₁+ T₂ - T₁/ 8₂ - 8₁ ( s - s₁)

= (400 + 500 - 400/8.3271 - 8.0347 (8.2652 - 8))

= 478.83°C

The final enthalpy is determined in the same way:

h₂= h₁ + h₂ - h₁/s₂ - s₁ ( s - s₁)

= ( 3275.5+ 3486.6 - 3275.5/ 8.3271 - 8.0347) (8.265)

= 3441.91 kJ/kg

6 0

2 years ago

2.

schepotkina [342]

Answer:

b. 6.02 x 1023 molecules

Explanation:

The formula mass of ammonia is 14 + 1 × 3 = 17.

The number of moles in 27.6g ammonia is 27.6 ÷ 17 = 1.62 mol.

A mole is 6.02 × 10²³, so the number of hydrogen atoms in a 1.62 moles of ammonia is 1.62 × 6.02 × 10²³ × 3 = 2.93 × 10² atoms.

3 0

3 years ago

What is the mass of oxygen in 3.34 g of potassium permanganate?

3241004551 [841]

Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

Percent of Oxygen (O) = 64 / 158 x 100

Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

mass of Oxygen (O) = 0.405 x 3.34 g

mass of Oxygen (O) = 1.35 g

5 0

4 years ago

If two pretty beat friends exist would they actually be real or just 2 dogs1.yes2.no