Answer:
158.5g Zn are produced
Explanation:
To solve this question we have to find the moles of Aluminium. With the moles of Aluminium and the balanced reaction we can find the moles of Zn and its mass as follows:
<em>Moles Al -Molar mass: 26.98g/mol</em>
43.6g Al* (1mol/26.98g) = 1.616 moles Al
<em>Moles Zn:</em>
1.616 moles Al * (3mol Zn / 2mol Al) =
2.424 moles Zn are produced
<em>Mass Zn -Molar mass: 65.38g/mol-</em>
2.424 moles Zn * (65.38g / mol) =
<h3>158.5g Zn are produced</h3>
Answer:
d. they are both involved in the division of somatic cells
Explanation:
Mitosis and meiosis are both types of cell divisions that occur in the system of living organisms. However, mitosis and meiosis are different in the type and number of daughter cells they give rise to and also where they occur.
Mitosis, which takes place in the SOMATIC OR BODY CELLS of living organisms, give rise to two daughter cells that are genetically identical to their parent cell. On the other hand, meiosis takes place strictly in the CELLS OF REPRODUCTIVE ORGANS, and gives rise to four genetically different daughter cells.
<span> .789=12.8*x
.789/12.8=x
.0616=x
.0616ml </span>
Answer:
There are 2.258 x 10^24 molecules of CO2 in 165 grams of CO2.
Explanation:
First you have to calculate the molar mass of carbon dioxide ⇒ 44 grs/mol. Then,
44 grs ------------- 1 mol CO2
165 grs ------------ x = 3.75 moles of CO2.
Then, from Avogadros constant we have that there are 6.022 x 10^23 molecules in 1 mol. So,
1 mol CO2 ---------------- 6.022 x 10^23 molecules CO2
3.75 moles CO2-------- x = 2.258 x 10^24 molecules CO2.
There are 2.258 x 10^24 molecules CO2 in 165 grams of CO2.
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
