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2 years ago

Calculate density of aluminum if 27.6cm^3 has a mass of 74.6g

Chemistry

2 answers:

rjkz [21]2 years ago

6 0

First , 27.6 cm^3 is equal to 27.6 ml

Density = mass g / volume cm^3

= 74.6 / 27.6 = ........ g/cm^3

PolarNik [594]2 years ago

5 0

Answer : The density of aluminium will be, $2.73g/cm^3$

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

$Density=\frac{Mass}{Volume}$

Given :

Mass of aluminum = 74.6 grams

Volume = $27.3cm^3$

Now put all the given values in the above formula, we get the density of aluminium.

$Density=\frac{74.6g}{27.3cm^3}=2.73g/cm^3$

Therefore, the density of aluminium will be, $2.73g/cm^3$

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A gaseous mixture of O2 and N2 contains 37.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the tot

kondor19780726 [428]

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

$p_A=\chi_A\times P_T$

$p_{O_2}$ = partial pressure of $O_2$ = ?

$\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}$

$P_{T}$ = total pressure of mixture = 525 mmHg

${\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles$

${\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles$

Total moles = 1.94 + 1.35 = 3.29 moles

$\chi_{O_2}=\frac{1.94}{3.29}=0.59$

$p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg$

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

7 0

3 years ago

Calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) where ΔH∘ = -687 kJ and ΔS∘ = -169 J/

aivan3 [116]

Answer:

-0.129V

Explanation:

The change in free energy is obtained from the given parameters after which the value is now applied to obtain the cell potential in volts from the formukar shown in the solution below.

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3 years ago

If all other variables remain unchanged, what happens to the output force when the area of the input piston is doubled?

mars1129 [50]

Answer:

1250N

Explanation:

This question is based on pascal's Law.

So By Pascal's Law

therefore =force on input piston =25N

= Force or weight on output person.

therefore after putting the values we get,

= (25x 1500)/30

=1250N

3 0

3 years ago

Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete

oee [108]

Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.

5 0

3 years ago

One molecule of glucose makes 36 molecules of atp how many molecules of glucose are needed to make 1800 molecules of atp

pogonyaev

You would need exactly 50 molecules of glucose.

3 0

3 years ago