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4 months ago

A person loses 2.70 lbs in two weeks how many grams did he lose?

Chemistry

2 answers:

VMariaS [2.9K]4 months ago

6 0

He lost 1224.7 grams.

Alekssandra [3K]4 months ago

5 0

I concur with angeliticsoul; the individual lost 1224.7 grams.

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It is 762 miles from here to Chicago. An obese physics teacher jogs at a rate of 5.0 miles every 20.0 minutes. How long would it

Anarel [2989]

3,048 minutes. Explanation: 762 divided by 5, then multiply that number by 20.

5 0

2 months ago

If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr

VMariaS [2998]

Answer:

The heat capacity of the calorimeter is $C_c$ = 54.4 $\frac{J}{c}$

Explanation:

Given the data

Heat supplied Q = 4.168 KJ = 4168 J

Mass of water $m_w$ = 75.40 gm

Change in temperature = ΔT = 35.82 - 24.58 = 11.24 °C

From the conditions provided

Q = $m_w C_w$ ΔT + $C_c$ ΔT

Plugging all values into the above equation yields

4168 = 75.70 × 4.18 × 11.24 + $C_c$ × 11.24

611.37 = $C_c$ × 11.24

$C_c$ = 54.4 $\frac{J}{c}$

This represents the heat capacity of the calorimeter.

7 0

3 months ago

What volume of gold would be equal in mass to a piece of copper with a volume of 141 ml? the density of gold is 19.3 g/ml; the d

alisha [2963]

We need to calculate the volume of Gold, assuming its mass matches that of copper.

Given information:

Density of Copper = 8.96 g/ml.
Volume of Copper = 141 ml.
Mass of Gold = Mass of Copper.
Density of Gold = 19.3 g/ml.

To find copper's mass, we use the density equation:
Density = mass/volume.

To find mass of copper:
Mass of copper = Density of Copper * Volume of Copper.
Mass of copper = 8.96 g/ml * 141 ml = 1263.36 g.
Thus,
Mass of gold = Mass of copper = 1263.36 g.
Now, using the density formula for gold to get its volume:
Volume of gold = Mass of gold / Density of gold.
Volume of gold = 1263.36 g / 19.3 g/ml = 65.46 mL.

Consequently, the volume of gold required to match the mass of copper is 65.46 mL.

8 0

3 months ago

Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol. Express your a

castortr0y [3046]

Answer:

A) The molecular formula for ibuprofen is $C_{13}H_{18}O_2$

B) The molecular formula for Cadaverine is $C_{5}H_{14}N_2$

C) The molecular formula for Epinephrine is $C_9H_{13}O_3N_1$

Explanation:

Element percentage in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100$

A) The composition of ibuprofen, used for headaches, consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by weight.

Ibuprofen has a molar mass of 206 g/mol.

The proposed molecular formula for ibuprofen is = $C_xH_yO_z$

Count of carbon atoms in one ibuprofen molecule;

$75.69\%=\frac{x\times 12 g/mol}{206 g/mol}\times 100$

$x=\frac{75.69\times 206 g/mol}{100\times 12 g/mol}=12.99\approx 13$

Count of hydrogen atoms in one ibuprofen molecule;

$8.80\%=\frac{y\times 1 g/mol}{206 g/mol}\times 100$

$y=\frac{8.80\times 206 g/mol}{100\times 1 g/mol}=18.12\approx 18$

Count of oxygen atoms in one ibuprofen molecule;

$15.51\%=\frac{z\times 16 g/mol}{206 g/mol}\times 100$

$z=\frac{15.51\times 206 g/mol}{100\times 16 g/mol}=1.99\approx 2$

Molecular formula for ibuprofen:

= $C_xH_yO_z= C_{13}H_{18}O_2$

B) Cadaverine consists of 58.55% carbon, 13.81% hydrogen, and 27.40% nitrogen by weight

Cadaverine has a molar mass of 102.2 g/mol.

The proposed molecular formula for Cadaverine is = $C_xH_yN_z$

Count of carbon atoms in one Cadaverine molecule;

$58.55\%=\frac{x\times 12 g/mol}{102.2 g/mol}\times 100$

$x=\frac{58.55\times 102.2 g/mol}{100\times 12 g/mol}=4.98\approx 5$

Count of hydrogen atoms in one Cadaverine molecule;

$13.81\%=\frac{y\times 1 g/mol}{102.2 g/mol}\times 100$

$y=\frac{13.81\times 102.2 g/mol}{100\times 1 g/mol}=14.11\approx 14$

Count of nitrogen atoms in one Cadaverine molecule;

$27.40\%=\frac{z\times 14 g/mol}{102.2 g/mol}\times 100$

$z=\frac{27.40\times 102.2 g/mol}{100\times 14 g/mol}=2.00\approx 2$

Molecular formula for Cadaverine:

= $C_xH_yN_z= C_{5}H_{14}N_2$

C) Epinephrine includes 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by weight

Epinephrine has a molar mass of 180 g/mol.

The proposed molecular formula for Epinephrine is = $C_xH_yO_zN_w$

Count of carbon atoms in one Epinephrine molecule;

$59.0\%=\frac{x\times 12 g/mol}{180 g/mol}\times 100$

$x=\frac{59.0\times 180 g/mol}{100\times 12 g/mol}=8.85\approx 9$

Count of hydrogen atoms in one Epinephrine molecule;

$7.1\%=\frac{y\times 1 g/mol}{180 g/mol}\times 100$

$y=\frac{7.1\times 180 g/mol}{100\times 1 g/mol}=12.78\approx 13$

Count of oxygen atoms in one Epinephrine molecule;

$26.2\%=\frac{z\times 16 g/mol}{180 g/mol}\times 100$

$z=\frac{26.2\times 180 g/mol}{100\times 16 g/mol}=2.94\approx 3$

Count of nitrogen atoms in one Epinephrine molecule;

$7.7\%=\frac{w\times 14 g/mol}{180 g/mol}\times 100$

$w=\frac{7.7\times 180 g/mol}{100\times 14 g/mol}=0.99\approx 1$

Molecular formula for Epinephrine:

= $C_xH_yO_zN_w= C_9H_{13}O_3N_1$

7 0

4 months ago