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creativ13 [48]
3 years ago
14

Mg3N2+3H2O-3MgO+2NH3 If 17 g of magnesium nitride is used, what volume of ammonia gas would be collected at 20°C and 0.989 atm?​

Chemistry
1 answer:
12345 [234]3 years ago
8 0

Explanation:

what are the unit given for the questions

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The ion with 36 electrons, 34 protons, and 44 neutrons
Reil [10]
While the number nuclear protons as given is 34, and therefore we deal with the element selenium, there are 2 more electrons than protons, and therefore this species has an overall
2
−
charge.
We represent this selenide ion as
S
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2
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. Do I win 5 pounds?

Z= 34, therefore the atom is selenium
7 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
What types of atoms will experience radioactivity?
Otrada [13]

Answer:

Alpha particles, Beta particles, Gamma rays

Alpha Is the least harmful

Beta is second most harmful

Gamma is most dangerous

Explanation:

3 0
3 years ago
What are the missing coefficents for the following equation? __N2+__H2-&gt;__NH3
vivado [14]

Answer:

1, 3, 2

Explanation:

N2 + H2 → NH3

I usually find that the best way to systematically balance an equation by inspection is to start with the most complicated-looking formula and then balance atoms in the order:

  1. All atoms other than O and H
  2. O
  3. H

(a) The most complicated formula is NH3.

(b) Balance N.

We have 1 H in NH3, but 2 N on the left. We need 2 N on the right. Put a 1 in front of N2 and a 2 in front of NH3.

1N2 + H2 → 2NH3

(c) Balance H.

We have fixed 6 H on the right, so we need 6 H on the left. Put a 3 in front of H2.

1N2 + 3H2 → 2NH3

The equation is now balanced, and the coefficients are 1, 3, 2.

5 0
3 years ago
Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it
Sholpan [36]

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

t__{1/2}}= \frac{0.693}{k}

where;

k = constant

The formula can be re-written as:

k = \frac{0.693}{t__{1/2}}

k = \frac{0.693}{9.0 min}

k = 0.077 min^{-1}

Let the initial amount of butter flavor in the food be (N_0) = 100%

Also, the amount of butter flavor retained at 200°C (N_t)= 74%

The rate constant k = 0.077 min^{-1}

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

\frac{N_t}{N_0}= -kt

t = - (\frac{1}{k}*In\frac{N_t}{N_0}  )

Substituting our values; we have:

t = - (\frac{1}{0.077}*In\frac{74}{100}  )

t = 3.91 minutes

∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes

4 0
3 years ago
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