The atom has only one isotope which means 100 % of same atom is present in nature. The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. Mass of one atom of that isotope is 9.123 ✕ 10⁻²³ g, so mass of one mole of atom that is Avogadro's number of atom is 6.023 X 10²³ X 9.123 X 10⁻²³ g=54.94 g = 55 g (approximate).
So, the atom having atomic mass 55 will be Cesium (Cs). Only one isotope of Cesium is stable in nature.
Answer:
The answer to be filled in the respective blanks in question is
3 and 1
Explanation:
So, we know that the formation of cabon-dioxide mole and that of Adenosin-Tri-Phosphate (ATP) moles will be in the ratio of 3:1 i.e., three carbon-di-oxide moles and 1 ATP mole.
Therefore, we can say that one pyruvate mole when passed through citric acid cycle and pyruvate dehydrogenase yields carbon-di-oxide and ATP moles in the ratio 3:1
Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
Answer:
Autotrophs are known as producers because they are able to make their own food from raw materials and energy. Examples include plants, algae, and some types of bacteria. Heterotrophs are known as consumers because they consume producers or other consumers. Dogs, birds, fish, and humans are all examples of heterotrophs
Explanation:
So they are heterotrophs
Ticks are tiny parasites that feed on the blood of their hosts (humans and animals) in order to survive and advance to the next life cycle stage.
Answer:
See explanation below for answers
Explanation:
We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.
a) mass of displaced water.
In this case all we need to do is to substract the 0.70 with the 0.13 g. so:
mW = 0.70 - 0.13
mW = 0.57 g of water
b) Volume of water.
In this case, we have the density of water, so we use the formula for density and solve for volume:
d = m/V
V = m/d
Replacing:
Vw = 0.57/0.9982
Vw = 0.5710 mL of water
c) volume of the metal weight
In this case the volume would be the volume displaced of water, which would be 0.5710 mL
d) the mass of the metal weight.
In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g
e) density.
using the above formula of density we calculate the density of the metal
d = 0.70 / 0.5710
d = 1.2259 g/mL
