Typically, whole blood contains all of the following types of blood cells EXCEPT:

1 answer:

7 0

That would be C because sickle cells make you very sick and could lead to death and I remember it by thinking sickle has the word sick in it.

You might be interested in

One of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF7). How many moles of CsXeF7 can be produced from t

larisa [96]

The balanced chemical equation is written as:

CsF(s) + XeF6(s) ------> CsXeF7(s)

We are given the amount of cesium fluoride and xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:

11.0 mol CsF ( 1 mol CsXeF7 / 1 mol CsF ) = 11.0 mol CsXeF7

6 0

3 years ago

Read 2 more answers

Consider the generic reaction: 2 A(g) + B(g) → 2 C(g). If a flask initially contains 1.0 atm of A and 1.0 atm of B, what is the

irina1246 [14]

Answer:

b. 1.5 atm.

Explanation:

Hello!

In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:

$P=P_A+P_B+P_C$

Now, if we introduce the stoichiometry, and the change in the pressure $x$ we can write:

$P=1.0-2x+1.0-x+2x$

Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:

$x=\frac{1.0atm}{2}=0.5atm$

Thus, the final pressure is:

$P=1.0-2(0.5)+1.0-(0.5)+2(0.5)\\\\P=1.5atm$

Therefore the answer is b. 1.5 atm.

Best regards!

3 0

3 years ago

What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer

Zepler [3.9K]

$d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}$