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3 years ago

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A balloon at 32 °C is filled with 21 L of air. What would its volume be at a temperature of 52 °C, assuming pressure remains con

stant?
A.)19 L
B.)22 L
C.)24 L
D.)25 L

1 answer:

I am Lyosha [343]3 years ago

4 0

Answer:C). 24L

Explanation: it works

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Calculate OH- given H3 O+ = 1.40 *10^-4 M

vova2212 [387]

3.16 X 10^-11 M is the [OH-] concentration when H3O+ = 1.40 *10^-4 M.

Explanation:

data given:

H30+= 1.40 X 10^-4 M\

Henderson Hasslebalch equation to calculate pH=

pH = -log10(H30+)

putting the values in the equation:

pH = -log 10(1.40 X 10^-4 M)

pH = 3.85

pH + pOH =14

pOH = 14 - 3.85

pOH = 10.15

The OH- concentration from the pOH by the equation:

pOH = -log10[OH-]

10.5= -log10[OH-]

[OH-] = 10^-10.5

[OH-] = 3.16 X 10^-11 is the concentration of OH ions when hydronium ion concentration is 1.40 *10^-4 M.

8 0

3 years ago

A substance is found to consist of 1.900g of silver, 0.250g of nitrogen, and 0.850g of oxygen. what is the percent by mass of ox

Galina-37 [17]

The percentage by mass of oxygen in the compound

find the total mass=( 1.900+ 0250 +0.850) = 3
the percentage mass mass of oxgyen/total mass x100

that is (0.850/3) x100=28.33%

7 0

3 years ago

X(g)+4Y(g)→2Z(g), ΔH∘=−75.0 kJ Before the reaction, the volume of the gaseous mixture was 5.00 L. After the reaction, the volume

vlabodo [156]

Isobaric transition, first law: <span>H=ΔU+w</span> for a gas expansion: <span>w=<span>P<span>ext</span></span>∗ΔV</span> to convert to joules, you need the gas constants. R = 0.08206 L atm/mol*K, R=8.314 J/mol*K <span>w=<span>P<span>ext</span></span>∗ΔV∗<span><span>8.314 J/mol∗K</span><span>0.08206 L atm/mol∗K</span></span></span> <span>ΔU=ΔH−[<span>P<span>ext</span></span>∗ΔV∗<span><span>8.314 J/mol∗K</span><span>0.08206 L atm/mol∗K</span></span>]</span> <span>ΔU=−75000 J−[(43.0atm)∗(2−5)L∗<span><span>8.314 J</span><span>0.08206 L atm</span></span>]</span> Then you need to convert to kJ. by the way U=E, internal energy.

7 0

3 years ago

Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o

Nata [24]

<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M$

where,

$n_p$ = number of protons

$m_p$ = mass of one proton

$n_n$ = number of neutrons

$m_n$ = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is $_{15}^{31}\textrm{P}$

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

$\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399$

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

8 0

3 years ago

Hydrogen iodide decomposes slowly to H2 and I2 at 600 K. The reaction is second order in HI, and the rate constant is 9.7×10−6M−

Lady bird [3.3K]

Answer : The molarity after a reaction time of 5.00 days is, 0.109 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $9.7\times 10^{-6}M^{-1}s^{-1}$

t = time taken = 5.00 days

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.110 M

Now put all the given values in above equation, we get:

$9.7\times 10^{-6}=\frac{1}{5.00}\left (\frac{1}{[A]}-\frac{1}{(0.110)}\right)$

$[A]=0.109M$

Hence, the molarity after a reaction time of 5.00 days is, 0.109 M

8 0

3 years ago