Which has a higher freezing point, pure water or sea water?

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g

satela [25.4K]

Answer:

$5.6gNa_2SO_4$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

$n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4$

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

$m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4$

Best regards.

5 0

2 years ago

A company is testing drinking water and wants to ensure that ca content is below 155 ppm. What is the maximum amount of Ca that

BabaBlast [244]

Answer:

138 mg

Explanation:

A company is testing drinking water and wants to ensure that Ca content is below 155 ppm (= 155 mg/kg), that is, <em>155 milligrams of calcium per kilogram of drinking water</em>. We need to find the maximum amount of calcium in 890 g of drinking water.

Step 1: Convert the mass of drinking water to kilograms.

We will use the relation 1 kg = 1000 g.

$890g \times \frac{1kg}{1000g} =0.890kg$

Step 2: Calculate the maximum amount of calcium in 0.890 kg of drinking water

$0.890gH_2O \times \frac{155mgCa}{1kgH_2O} = 138mgCa$

7 0

3 years ago

In a certain city, electricity costs $0.17 per kW·h. What is the annual cost for electricity to power a lamp-post for 5.50 hours

Anon25 [30]

Answer:

(a) = $34.123

(b) = $8.532

(c) Additional cost of fluorescent bulb is justified

Explanation:

Cost of electricity = $0.17 per kW·h

(a) For a 100 watt bulb which is the same as 100/1000 or 0.1 kW, the cost per hour =

0.1 × 0.17 = $0.017/h

and for 5.5 hours = 0.017×5.5 = $0.0935

The annual cost, which is 365 days, we have

Annual cost = $0.0935 × 365 = $34.123

(b) For the energy efficient 25-watt bulb, we have

25/1000 = 0.025kW

Power cost per annum =

0.025kW×$0.17 per kW·h×5.5×365 = $8.532

(c) Total cost of incandescent bulb = $0.89 total cost of using the incandescent bulb is $34.123 + $0.89 = $35.02

Total cost of using the energy efficient fluorescent bulb is about $3.49

Total cost of using the energy efficient bulb = $8.532 + $3.49 = $12.02

Total cost of incandescent bulb = $35.02 while total cost of energy efficient bulb is = $12.02

$12.02 <$35.02

Additional cost of fluorescent bulb is justified

8 0

3 years ago

Which of the following is true for the actual yield of a reaction?

ludmilkaskok [199]

Answer:

It is always less than the theoretical yield

Explanation:

For many chemical reactions, the actual yield is usually less than the theoretical yield. This is due to possible loss in the process or inefficiency of the chemical reaction.

4 0

2 years ago

Read 2 more answers

Please match the example with the classification of matter in which it belongs. Column A 1. soil: soil 2. chocolate milk: chocol

bekas [8.4K]

Answer:

34

Explanation:

add

8 0

2 years ago