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3 years ago

What is the mass of the polypropylene in grams? Assume that 1 lbs = 453.952 grams. Drag and drop the

Chemistry

1 answer:

Rudik [331]3 years ago

7 0

Answer:

7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs

7.50×10¯² lbs = 34.0464 g.

Explanation:

From the question given, the following data were obtained:

1 lbs = 453.952 g

7.50×10¯² lbs =.?

Thus, we can obtain convert 7.50×10¯² lbs to grams as follow:

1 lbs = 453.952 g

Therefore,

7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs

7.50×10¯² lbs = 34.0464 g

Therefore, 7.50×10¯² lbs is equivalent to 34.0464 g.

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A metal sample weighing 129.00 grams and at a temperature of 97.8 degrees Celsius was placed in 45.00 grams of water in a calori

Nitella [24]

Answer : The specific heat of metal is $0.481J/g^oC$ .

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of metal = 129.00 g

$m_2$ = mass of water = 45.00 g

$T_f$ = final temperature = $39.6^oC$

$T_1$ = initial temperature of metal = $97.8^oC$

$T_2$ = initial temperature of water = $20.4^oC$

Now put all the given values in the above formula, we get

$129.00g\times c_1\times (39.6-97.8)^oC=-45.00g\times 4.184J/g^oC\times (39.6-20.4)^oC$

$c_1=0.481J/g^oC$

Therefore, the specific heat of metal is $0.481J/g^oC$ .

4 0

3 years ago

What element is 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f10

skad [1K]

Answer:

Bi (Bismuth)

4 0

3 years ago

Read 2 more answers

5.6 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of

tresset_1 [31]

Answer:

0.78 atm

Explanation:

Step 1:

Data obtained from the question. This includes:

Mass of CO2 = 5.6g

Volume (V) = 4L

Temperature (T) =300K

Pressure (P) =?

Step 2:

Determination of the number of mole of CO2.

This is illustrated below:

Mass of CO2 = 5.6g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Number of mole CO2 =?

Number of mole = Mass/Molar Mass

Number of mole of CO2 = 5.6/44

Number of mole of CO2 = 0.127 mole

Step 3:

Determination of the pressure in the container.

The pressure in the container can be obtained by applying the ideal gas equation as follow:

PV = nRT

The gas constant (R) = 0.082atm.L/Kmol

The number of mole (n) = 0.127 mole

P x 4 = 0.127 x 0.082 x 300

Divide both side by 4

P = (0.127 x 0.082 x 300) /4

P = 0.78 atm

Therefore, the pressure in the container is

3 0

3 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

slava [35]

Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

8 0

2 years ago

State 3 differences between a solution, suspension and colloid.

Art [367]

Explanation:

solution 1) homogeneous

2) do not scatter light

3) cannot be separated by filtration

Colloids 1) heterogeneous

2) scatter light

3) cannot be separated by filtration

suspension 1) heterogeneous

2) may either scatter light or be opaque

3) can be separated by filtration

8 0

3 years ago