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2 years ago

This same chemistry student has a weight of 155 lbs. What is the student’s weight in

Chemistry

2 answers:

Virty [35]2 years ago

8 0

Answer:

the same chemistry student has a weight of 155lbs what is the student weight in grams? (16 oz= 1 lb, 1 oz= 28.34g)

Explanation:1 lb = 16oz, so multiply your pounds by 16 to get you ounces of the student, then multiply by 28.34 to get grams

155 X 16 X 28.34 = 70283.2

FinnZ [79.3K]2 years ago

7 0

Answer:

The student weighs 70306.8 grams.

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the hydrogen gas generated when calcium metal reacts with water is collected over water at 20 degrees C. The volume of the gas i

Gala2k [10]

Answer:

There is 0.0677 grams of H2 gas obtained

Explanation:

Step 1: Data given

The total pressure (988 mmHg) is the sum of the pressure of the collected hydrogen + the vapor pressure of water (17.54 mmHg).

ptotal = p(H2)+ p(H2O)

p(H2) = ptotal - pH2O = 988 mmHg - 17.54 mmHg = 970.46 mmHg

Step 2: Calculate moles of H2 gas

Use the ideal gas law to calculate the moles of H2 gas

PV = nRT

n = PV / RT

⇒ with p = pressure of H2 in atm = 970.46 mmHg * (1 atm /760 mmHg) = 1.277 atm

⇒ V = volume of H2 in L = 641 mL x (1 L / 1000 mL) = 0.641 L

⇒ n = the number of moles of H2 = TO BE DETERMINED

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

⇒ T = the temperature = 20.0 °C = 293.15 Kelvin

n = (1.277)(0.641) / (0.08206)(298.15) = 0.0335 moles H2

Step 3: Calculate mass of H2

Mass of H2 = moles H2 ¨molar H2

0.0335 moles H2 * 2.02 g/mol H2 = 0.0677g H2

There is 0.0677 grams of H2 gas obtained

7 0

2 years ago

A 0.530 M Ca(OH)2 solution was prepared by dissolving 36.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol

Paladinen [302]

The concentration of a solution is the number of moles of solute per fixed volume of solution.

Concentration (C) = number of moles of solute (n) / volume of the solution (v)

we have to find the volume of the solution when 36.0 g of Ca(OH)₂ is added to water to make a solution of concentration 0.530 M

mass of Ca(OH)₂ added - 36.0 g

number of moles of Ca(OH)₂ - 36.0 g / 74.1 g/mol = 0.486 mol

we know the concentration of the solution prepared and the number of moles of Ca(OH)₂ added, substituting these values in the above equation, we can find the volume of the solution

C = n/v

0.530 mol/L = 0.486 mol / V

V = 0.917 L

answer is 0.917 L

6 0

2 years ago

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Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = 19,26 kJ

I hope it helps!

5 0

2 years ago

The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the or

Aloiza [94]

Answer:

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life Carbon 14 = 5730 year

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ hour^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}$

$\frac {[A_t]}{[A_0]}=0.007015$

In percentage, the sample of C-4 remains = 0.7015 %

3 0

2 years ago

The graph below shows the electronegativities of the elements in the periodic table. A graph with the horizontal axis labeled pe

kow [346]

Answer:

The elements in group one

Explanation:

8 0

2 years ago

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