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3 years ago

If Uranium-240 emits an alpha particle, what does it transmutate into?

Chemistry

1 answer:

aliina [53]3 years ago

7 0

Answer:

i think is b but im not sure

Explanation:

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Law Incorporation [45]

Answer:

what about just talk and become friends :) :) :)

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3 years ago

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Most of the sulfur used in the United States is chemically synthesized from hydrogen sulfide gas recovered from natural gas well

Rom4ik [11]

Answer:

The rate at which sulfur dioxide is being produced is 0.90 kg/s.

Explanation:

Volume of oxygen gas consumed in second ,V= 994 L

Pressure of the gas = p

Temperature of the gas = T = 170°C= 170 + 273 K=443 K

Moles of oxygen gas consumed in a second = n

$PV=nRT$ ( ideapl gas equation)

$n=\frac{PV}{RT}=\frac{0.77 atm\times 994 L}{0.0821 atm L/mol K\times 443 K}$

n = 21.044 mole

Moles of dioxygen gas consumed per second = 21.044 mol

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$ (Claus process)

According to reaction, 3 moles of dioxygen gives 2 moles of sulfur dioxide gas.Then 21.044 moles of dioxygen will give;

$\frac{2}{3}\times 21.044 ol=14.029 mol$ of sulfur dioxide

Mass of 14.029 moles of sulfur dioxide gas;

14.029 mol × 64 g/mol = 897.86 g

897.86 g = 0.89786 kg ≈ 0.90 kg

Mass of sulfur dioxide produced per second = 0.90 kg

The rate at which sulfur dioxide is being produced is 0.90 kg/s.

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4 years ago

Drag the handle on the left and adjust the length of the box to 5 nm. (Because no other dimensions of the box change, its length

kvv77 [185]

Answer:

Length (nm) Pressure (atm)

5.0 11.7

6.0 9.8

7.0 8.4

8.0 7.2

9.0 6.6

10.0 5.8

Explanation:

This is also PLATOS answer!!

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3 years ago

Please tell me the answer I will mark you brainliest so please

s344n2d4d5 [400]

Answer: a animal cell

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3 years ago

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At what substrate concentration would an enzyme with a K_cat of 30.0 s⁻¹ and a Km of 0.0050 M operate at one-quarter of its maxi

Umnica [9.8K]

Answer:

The correct answer is option B.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}$

$V_{max}=k_{cat}[E_o]$

v = rate of formation of products

[S] = Concatenation of substrate = ?

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$E_o$ = initial concentration of enzyme

We have :

$v=\frac{V_{max}}{4}$

[S] =?

$K_m=0.0050 M$

$v=V_{max}\times \frac{[S]}{(K_m+[S])}$

$\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}$

$[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M$

So, the correct answer is option B.

8 0

3 years ago