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2 years ago

9

A 36-gram sample of water has an initial temperature of 22°c. After the sample absorbs 1200 joules of heat energy, the final tem

perature of the sample is.

1 answer:

professor190 [17]2 years ago

7 0

Answer:

≈29.94 [°C].

Explanation:

all the details are in the attachment, the answer is underlined with orange colour.

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Which of the following statements is not correct?

velikii [3]

To answer the question above, multiply the given number of moles by the molar masses.

(A) (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B) (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C) (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D) (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O

The answer to the question above is letter D.

4 0

4 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

Hello, I study Home MS Physical Science B-IC. Are there quizlets I can look up for the cumulative exam I have to take today?

Ludmilka [50]

Yes if you search up your subject or topic then put quizlet you’ll find your answer but you may need to login in to get the best experience of studying that you want

8 0

3 years ago

What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?

Vsevolod [243]

The dilution formula can be used to find the volume needed

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

c1 - 0.33 M

c2 - 0.025 M

v2 - 25 mL

Substituting these values in the equation

0.33 M x v1 = 0.025 M x 25 mL

v1 = 1.89 mL

Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution

7 0

3 years ago

The elements in Groups 1A(1) and 7A(17) are all quite reactive. What is a major difference between them?

matrenka [14]

The elements in Groups 1A(1) and 7A(17) are all quite reactive.

<h3>Major difference between Groups 1A(1) and 7A(17) : </h3>

Group 7's halogens, which are non-metal elements, become less reactive as you move down the group. In contrast to the alkali metals in Group 1 of the periodic table, this trend is the opposite. The most reactive element in Group 7 is fluorine.

Alkali metals are soft and reactive metals. They react vigorously with water and become more reactive. And other hand halogens are reactive non metals.

Elements of group 1A are known as alkali metals. Elements of this group are lithium, sodium, potassium, rubidium, cesium.
Reactivity increase down group 1 but decrease up group 7 this is because group 7 elements react by gaining an electron. As one move down the group, the amount of electron shielding increases, meaning that the electron is less attracted to the nucleus.

To know more about Groups 1A(1) and 7A(17) please click here :

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6 0

2 years ago