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statuscvo [17]
3 years ago
15

Liquid nitrogen boils at –195.8°C. Express the boiling point of liquid nitrogen in kelvin

Chemistry
1 answer:
inessss [21]3 years ago
4 0

Answer:

\boxed {\boxed {\sf 77.35 \ K}}

Explanation:

The Celsius and Kelvin scales are used to measure the temperature of matter.  Their scales and unit differences are the same (1 K increase = 1 °C increase), but they have different starting points.

So, the conversion is quite simple and only requires addition because of the different starting points. The formula is:

T_K = T_C+ 273.15

The boiling point of liquid nitrogen is -195.8 °C. We can substitute this value into the formula.

T_K= -195.8 + 273.15

T_K= 77.35 K

The boiling point of liquid nitrogen is 77.35 Kelvin.

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After dipping the litmus paper strips in the solutions, the answer to the question is that one is acidic because according to brad's observation on turned red and the others are basic or alkaline with the two papers turning blue.
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3 years ago
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Using the balanced equation N2+O2=2NO, how many grams of NO can be produced when 25.0 grams of N react?
Oxana [17]

Answer:

53.55gNO

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:

=25.0gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNO}{1molN_2}*\frac{30.01 gNO}{1molNO}\\\\=53.55gNO

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8 0
3 years ago
If 9 moles of nitrogen gas and 9 moles of hydrogen
Luda [366]

Answer:

dnfberbgiieur

Explanation:

4 0
4 years ago
Suppose you have a 2.0 molar solution of sodium hydroxide (naoh), and you need 65 ml of 0.6 molar naoh. how should you make this
nalin [4]

The given molarity of sodium hydroxide solution = 2.0 M

The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH

Converting 65 mL to L:

65mL*\frac{1L}{1000mL} =0.065L

Calculating the moles of NaOH in the final solution:

0.065L * \frac{0.6 mol}{L} =0.039mol

Finding out the volume of 2.0 M solution taken to prepare the final solution:

0.039 mol * \frac{1L}{2.0mol}=0.0195L*\frac{1000mL}{1L} =19.5mL

Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.


3 0
3 years ago
Balancee las ecuaciones químicas por el método de tanteo e identifique que tipo de reacción es: 1.1.- Cloruro férrico acuoso rea
lakkis [162]

Respuesta:

2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)

Explicación:

Consideremos la ecuación no balanceada que ocurre cuando cloruro férrico acuoso reacciona con carbonato de sodio sólido para formar carbonato férrico sólido y cloruro de sodio acuoso. Esta es una reacción de doble desplazamiento.

FeCl₃(aq) + Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)

Vamos a usar el método de tanteo. Empezaremos balanceando los átomos de C, multiplicando Na₂CO₃ por 3.

FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)

Luego, balancearemos los átomos de Fe, multiplicando FeCl₃ por 2.

2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + NaCl(aq)

Finalmente, obtendremos la ecuación balanceada, multiplicando NaCl por 6.

2 FeCl₃(aq) + 3 Na₂CO₃(s) ⇒ Fe₂(CO₃)₃(s) + 6 NaCl(aq)

4 0
3 years ago
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