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ANTONII [103]
3 years ago
12

Select the number that are signicantbin the number 1500 ?

Chemistry
1 answer:
Andrew [12]3 years ago
8 0
1 and 5 are definitely significant. The zeros might not be significant .
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If I have 22,9 moles of a gas at a pressure of 12.4 atm and a volume of 24 liters, what is the temperature?
Grace [21]

Answer:

160 K, using 2 SF,

Explanation:

Ideal Gas law, rearrange to solve for T (K), answer about 158.3 round to 2SF

PV = nRT

8 0
2 years ago
What does mechanical energy mean?
Tamiku [17]

Answer:

Mechanical energy is basically the energy of movement

For example, if I am running, im using Mechanical energy

3 0
3 years ago
Read 2 more answers
25 points!! Will give Brainliest What is the molarity of the dilute solution if you dilute 35.0 mL of a 12.0 M solution to make
olga2289 [7]

Answer:

0.72 good luck with the answer

3 0
3 years ago
A presión de 3atm, 7L de un gas a temperatura constante experimenta un cambio ocupando un volumen de 0,5L ¿Cuál será la presión
pantera1 [17]

Answer:

La presión que ejerce es 42 atm.

Explanation:

La ley de Boyle relaciona la presión y el volumen y dice que  el volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión.

La ley de Boyle se expresa matemáticamente como:

P*V=k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1= P2*V2

En este caso:

  • P1= 3 atm
  • V1= 7 L
  • P2= ?
  • V2= 0.5 L

Reemplazando:

3 atm* 7 L= P2* 0.5 L

Resolviendo:

P2=\frac{3 atm*7 L}{0.5 L}

P2= 42 atm

<u><em>La presión que ejerce es 42 atm.</em></u>

3 0
3 years ago
A 100g of sample of a compound is combusted in excess oxygen and the products are 2.492g of CO2 and 0.6495 of H2O. Determine the
ruslelena [56]

The empirical formula : C₁₁O₁₄O₃

<h3>Further explanation</h3>

The assumption of the compound consists of C, H, and O

mass of C in CO₂ =

\tt \dfrac{12}{44}\times 2.492=0.680~g

mass of H in H₂O =

\tt \dfrac{2.1}{18}\times 0.6495=0.072~g

mass of O :

mass sample-(mass C + mass H)

\tt 1-(0.68+0.072)=0.248`g

mol of  C :

\tt \dfrac{0.68}{12}=0.056

mol of H :

\tt \dfrac{0.072}{1}=0.072

mol of O :

\tt \dfrac{0.248}{16}=0.0155

divide by 0.0155(the lowest ratio)

C : H : O ⇒

\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div  1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3

3 0
3 years ago
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