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2 years ago

Please help asap! Giving brainliest to correct answer!!

Chemistry

1 answer:

eduard2 years ago

3 0

The answer is d, because the simplest form of C4H10 is C2H5 which is the empirical formula

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A 0.1 mm sample of human blood has approximately 6000 red blood cells. An adult typically has 5.0 L of blood. How many red blood

olga55 [171]

3.0 × 10¹¹ RBC's (or) 3E11 RBC's

Solution:

Step 1: Convert mm³ into L;

As,

1 mm³ = 1.0 × 10⁻⁶ Liters

So,

0.1 mm³ = X Liters

Solving for X,

X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³

X = 1.0 × 10⁻⁷ Liters

Step 2: Calculate No. of RBC's in 5 Liter Blood:

As given

1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's

So,

5.0 Liters of Blood will contain = X RBC's

Solving for X,

X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters

X = 3.0 × 10¹¹ RBC's

Or,

X = 3E11 RBC's

5 0

2 years ago

Help me with this science question on the middle one..

viktelen [127]

Networks of feeding relationships is correct

8 0

2 years ago

What would the products of a double-replacement reaction between KBr and CaO be? (Remember: In double-replacement reactions, the

11111nata11111 [884]

ANSWER

$\begin{gathered} \text{ 2KBr }+\text{ CaO }\rightarrow\text{ K}_2O\text{ }+\text{ CaBr}_2 \\ Option\text{ B} \end{gathered}$

EXPLANATION

Given that;

The two reactants are KBr and CaO

Double replacement reaction is a type of chemical reaction that occur when two reactants exchange cations and anions to yield new products.

$\text{ 2KBr + CaO }\rightarrow\text{ K}_2O\text{ + CaBr}_2$

Therefore, the resulting products of the given data are K2O + CaBr2

The correct answer is option B

4 0

10 months ago

SOMEONE PLEASE HELPPP

xxTIMURxx [149]

A is the correct awnser Beacuse it ether right kne

5 0

2 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

Katen [24]

Answer:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

Explanation:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

$Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}$

The pKa is

$-Log (Ka) = -Log (6.2x10^{-8}) = 7.2$

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.

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3 years ago