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Lunna [17]
3 years ago
9

What is the volume of a substance with a mass of 64.9 g/ and a density of 2.1 g/ml?

Chemistry
1 answer:
maks197457 [2]3 years ago
4 0
It’s 34 I had this question
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Use a Punnett square to predict the cross of a homozygous tall parent with a homozygous short parent, if tall is dominant over s
ycow [4]

Answer:

All offspring are tall when a homozygous tall parent with homozygous short parent.

Explanation:

When we crossed  homozygous tall parent with homozygous short parent, we conclude that all offspring are tall, because homozygous short parent are supressed under the homozygous tall parent, due to law of dominance.

Law of dominance states that, recessive alleles are suppressed by dominant alleles but they can appear in F2 generation.

Using a punett square, we can predict the cross between homozygous tall and homozygous short parent.      

  The phenotypes are: All are tall plants (4:0).

5 0
3 years ago
Which of the following is not observed in a homologous series? ​
Stolb23 [73]

Answer:

Change in chemical properties

Explanation:

4 0
3 years ago
The amount of kinetic energy an object has depends on
Alex777 [14]
How much it has to drop and how heavy it is. Hope this is what you're looking for:)
3 0
3 years ago
Read 2 more answers
- La siguiente tabla, expone cómo desciende la temperatura (T) del aire con la altitud(h): Temperatura (ºC) 15 13,5 12 10,5 9 Al
Mrrafil [7]

Answer:

k = -0.006.

T₀ = 15 °C

Explanation:

Hola.

En este caso, considerando la gráfica mostrada en el archivo adjunto, podemos evidenciar que los datos dados se comportan de manera lineal, por lo que basado en la ecuación, T=k*h+To, podemos calcular la pendiente que basicamente es igual a k, tomando dos puntos en la gráfica:

k=\frac{12-13.5}{750-500}=-0.0006

Además, el valor de la temperatura inicial se puede extraer de la tabla, dado que esta es cuando la altura es 0 m, es decir 15 °C.

¡Saludos!

8 0
3 years ago
For lunch, a patient consumed 3 oz of skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk (see Table 3.
Volgvan

Lunch of a patient has 3 oz skinless chicken, 3 oz of broccoli, 1 medium apple, and 1 cup of nonfat milk

Energy content of 3 oz skinless chicken is = 110 kcal

Energy content of 3 oz broccoli = 30 kcal

Energy content of 1 medium apple = 60 kcal

Energy content of 1 cup non-fat milk = 90 kcal

So the kilocalories of energy patient obtained from lunch

                   = 110 kcal+ 30 kcal + 60 kcal + 90 kcal = 290 kcal


3 0
3 years ago
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