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2 years ago

3. If 175 grams of nitrogen reacted with hydrogen to make hydrogen, how much ammonia would be needed?

Chemistry

1 answer:

zmey [24]2 years ago

7 0

Answer:

37.5 g of H₂

Explanation:

N₂ (28g) + 3H₂ (6g) => 2NH₃ (34g)

Every 28g of N₂ needs 6g of H₂

=> Every 7g of N₂ needs 1.5g of H₂

=> 175g of N₂ needs 1.5 x 25 g of H₂

=> 37.5 g of H₂

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The compound potassium sulfide is a strong electrolyte. Write the transformation that occurs when solid potassium sulfide dissol

Alex17521 [72]

Answer:

K₂S(s) → 2K⁺(aq) + S²⁻(aq)

Explanation:

A strong electrolyte is a compound that, in solution, dissociates completely. For a compound XₐYₙ that dissolves in water:

XₐYₙ(s) → aXⁿ⁺(aq) + nYᵃ⁻(aq)

Now, potassium sulfide (K₂S), as a strong electrolyte dissolves in water thus:

K₂S(s) → 2K⁺(aq) + S²⁻(aq)

I hope it helps!

3 0

3 years ago

What is produced when an acid and an alkali react?

marta [7]

The answer is D: Salt and Water

6 0

3 years ago

Read 2 more answers

What do you think the volume of a prescription bottle would be? plz help me

GrogVix [38]

Answer: about 5 Mg

Explanation:

Because that sounds right

7 0

4 years ago

A chemist wishing to do an experiment requiring 47Ca2+(half-life = 4.536 days) needs 5.00 g of the nuclide. What mass of 47CaCO

Artyom0805 [142]

Answer: The answer is 6.78 grams.

Explanation: The equation used for solving this type of problems is:

$\frac{N}{N_0}=(\frac{1}{2})^n$

where, $N_0$ is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.

Number of half lives is calculated on dividing the given time by the half life.

n = time/half life

Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.

$4.536days(\frac{24hours}{1day})$

= 108.864 hours

So, $n=\frac{48.0}{108.864}$ = 0.441

Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:

$\frac{5.00}{N_0}=(\frac{1}{2})^0^.^4^4^1$

$\frac{5.00g}{N_0}=0.737$

$N_0=\frac{5.00g}{0.737}$

$N_0$ = 6.78 g

So, 6.78 g of the radioactive substance needs to be ordered.

4 0

4 years ago

Determine the mass of grams of led(II) sulfate that will dissolve in 2.50 x 10^2 ml water. Ksp= 1.8 x 10^-8

ki77a [65]

Answer:

9.86*10^(-3) g

Explanation:

PbSO4 ----> Pb^(2+) + SO4^(2-)

s s

Ksp = s²

s =√Ksp = √(1.8*10^-8) = 1.3*10^(-4) mol/L

The molar solubility PbSO4 = 1.3*10^(-4) mol/L.

2.50 *10^2 mL *1L/10³mL =0.250L

1.3*10^(-4)mol/L *0.250L*303.3 g/mol = 9.86*10^(-3) g

7 0

3 years ago