Un mol de un gas monoatómico cuyo calor absorbido es 40.0 J y cuyo trabajo recibido es 200.0 J experimenta un cambio de energía interna de 240.0 J.
Tenemos un mol de un gas monoatómico en un recipiente con un pistón y suceden las siguientes transferencias de energía:
- El gas absorbe 40.0 J de calor, ya que Q > 0.
- El gas recibe 200.0 J de trabajo, ya que Q > 0.
Podemos calcular el cambio de energía interna del gas (ΔU) usando la siguiente fórmula.
Un mol de un gas monoatómico cuyo calor absorbido es 40.0 J y cuyo trabajo recibido es 200.0 J experimenta un cambio de energía interna de 240.0 J.
Aprende mas: brainly.com/question/21913262
<em>7. Considere un sistema que contiene un mol de un gas monoatómico retenidito por un pistón. ¿Cuál es el cambio de energía interna del gas, Q: 40.0 J y W: 200.0J?</em>
Answer:
Membrane-bound organelles
Explanation:
Differential:
- Eukaryotic cells have membrane-bound organelles, while prokaryotic cells do not.
Similar:
- Both cells need nutrients, there are eukaryotic and prokaryotic organisms of algae.
- Both eukaryotic and prokaryotic cells have a plasma membrane.
- Both eukaryotic and prokaryotic cells desire nutrients.
Answer:
The correct answer is A.) during transcription
Explanation:
The pairing during DNA replication and transcription is a little bit different. During replication, A pairs with T, T with A, C with G, and G with C. In transcription T is replaced by U so when there is nucleotide adenine is present in the DNA sequence than uracil will be added against it as complementary nucleotide in mRNA during transcription.
So here 5'-GAT-3' nucleotide sequence is present in DNA and after transcription the nucleotide sequence will be 5'-CUA-5' because U is added against A during transcription. Therefore the correct answer is A.
Answer:
3/8
Explanation:
Martha has a widow's peak (dominant trait) and attached earlobes (recessive trait).
Martha's dad had a straight hairline (ww) and unattached earlobes (Ee, because she has the recessive alleles ee and both parents give us one allele).
This tells you that martha has a mother with at least one of both alleles dominant for widow peak and at least one recesive allele of attached earlobes
So, martha's alleles are: Ww and ee.
If she marries a man that is heterozygous for both traits (Ww and Ee) the probabilitys are
Ww Ee x Ww ee: WWEe, WwEe, wWee, and wwee
