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bezimeni [28]
3 years ago
9

What is the other name for white blood cells​

Chemistry
2 answers:
nlexa [21]3 years ago
7 0

Answer:

leukocytes

Explanation:

(We talked bout this in class yesterday)

zubka84 [21]3 years ago
5 0

Answer:

leukocyte

Explanation:

white blood corpuscle

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In the reaction Na2CO3 + 2HCl → 2NaCl + CO2 + H2O, how many grams of CO2 are produced when 7.5 moles of HCl is fully reacted?
Sidana [21]

Answer:

165 of CO₂.

Explanation:

In the reaction:

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

2 moles of HCl reacts producing 1 mole o CO₂

If 7.5 moles of HCl reacts, moles of CO₂ produced are:

7.5 moles of HCl ₓ ( 1 mol CO₂ / 2 mol HCl) = 3.75 mol CO₂. As molar mass of CO₂ is 44g/mol, mass of CO₂ is:

3.75 mol CO₂ ₓ (44g / 1mol) = <em>165 of CO₂ </em>

8 0
3 years ago
Read 2 more answers
The Same force is applied to a 300 kg go kart and 100 kg wagon at the beginning of the race. Which time is true about their acce
murzikaleks [220]

Answer: Well!

Explanation: I was going to answer D but fverdell82156 got to it first! So I have to agree with him! It is D!

3 0
3 years ago
I really need help on this question.
natita [175]
I need that answer too
7 0
3 years ago
Zinc oxide adopts a face-centered cubic arrangement. Both Zinc ions and oxide ions adopt the face-centered cubic arrangement; wi
vichka [17]

Answer:

5.41 ×10⁻²²

Explanation:

We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.

Then, the number of ZnO molecule in one unit cell = 4

The standard molar mass of ZnO = 81.38g

Avogadro's constant = 6.023 × 10²³ mole

∴

The mass of one unit cell of zinc oxide can be calculated as:

= \frac{4*81.38}{6.023*10^{23}}

= 5.40461564×10⁻²²

≅ 5.41 ×10⁻²²

∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²

3 0
3 years ago
Elements that exceed the octet rule must have an _______________. unoccupied s-orbital unoccupied p-orbital unoccupied d-orbital
Irina18 [472]
Elements that exceed the octet rule must have an unoccupied d orbital. I believe.
7 0
4 years ago
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