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DaniilM [7]
3 years ago
12

Can someone please help me on #17? I don’t get it :(

Chemistry
1 answer:
Dmitry [639]3 years ago
5 0
Given is the specific heat of water equal to 4.18 Joule per gram per *C.

This means to raise the temperature of 1 g of water by 1 degree Celsius we need 4.18 joule of energy.

Now, look at the question. We are asked that how much amount of energy would be required to raise the temperature of 25 g of water by (54-50) = 4 degree celsius.

To do so we have formula

Q = m C (temperature difference)

Have a look at pic for answer

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Calculate the amount of F.A.S required to prepare 1000 ml of 0.1 molar standard solution of F.A.S
RideAnS [48]

Answer:

Preparation and Standardization of 0.1 M Ferrous Ammonium...

Dissolve 40 g of ferrous ammonium sulfate in a previously cooled mixture of 40 ml of sulphuric acid and 200 ml of water.

Dilute with sufficient freshly boiled and cooled water to produce 1000 ml.

Standardize the solution in the following manner.

Explanation:

8 0
3 years ago
Why are marine mammals so important
sergejj [24]
Because they help regulate the marine environment.
5 0
3 years ago
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Why is the vapor pressure of a warm lake higher than the vapor pressure of a cold lake? A. Warm water has a greater heat of vapo
Naya [18.7K]

Answer : The correct option is B.

Explanation :

The temperature of the warm water is higher than the cold water. That means as the temperature of warm water increases, the kinetic energy of the warm water molecule also increases.

As the kinetic energy of the warm water molecule increases, the number of molecules conversion into a vapor also increases (i.e, quick evaporation), thereby increasing the vapor pressure of warm water.

Hence, the correct answer is, (B) Warm water evaporates more quickly.

7 0
3 years ago
Read 2 more answers
HS- is amphoteric; it can behave as either an acid or a base.
Tamiku [17]

Answer:

HS+Na=>NaS+1/2H2(here HS- acts as an acid)

HS-. + HCl=> H2S(g)+ Cl-(here HS- acts as a base)

7 0
3 years ago
In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The
Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

<u>Step 1: </u>Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

<u>Step 2:</u> Calculate mass of NaNO2

(40/100)*1.05  = 0.42 grams

<u>Step 3:</u> Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

<u>Step 4:</u> Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

<u>Step 5:</u> Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

<u>Step 6:</u> Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 =  0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

4 0
3 years ago
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