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3 years ago

11

PLS HELP FAST How does the circulatory system work at the tissue level? Cardiac muscle contracts. The heart pumps the blood. Nut

rients are delivered throughout the body. Blood cells carry nutrients.

2 answers:

Trava [24]3 years ago

8 0

Answer: it was B: the heart pumps the blood

Explanation:

i took the test and got an A

borishaifa [10]3 years ago

5 0

Answer:

b

Explanation:

You might be interested in

Why would a table be more suitable than a graph to show data

kow [346]

It might be more suitable because some data cannot use a graph to show data because there is too much data.

6 0

3 years ago

1. The volume of a gas is 10.0 liters and its temperature is 1500 K. If the temperature is

cupoosta [38]

Answer:

5 L

Explanation:

So this problem refers to Charles's law. You would use this formula..

$\frac{V_{I} }{T_{I} } = \frac{V_{F} }{T_{F} }$ (Initial volume / Initial Temperature) = (Final volume / Final Temperature)

For your problem, it would look like this...

$\frac{10L}{1500k} = \frac{V_{F} }{750k}$

You would cross multiply and your answer would be 5L.

To reassure yourself that the answer is correct, Charles law states that the Volume and the Temperature are directly proportional. Meaning if your temperature is decreasing, your volume <u>has</u> to decrease.

5 0

3 years ago

Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|

|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|

<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|

|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

3 years ago

Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.

Alborosie

<u>Given:</u>

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

<u>To determine:</u>

The Ka of HNO3

<u>Explanation:</u>

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

HNO3 ↔ H+ + NO3-

I 7.50 0 0

C -2.48 +2.48 +2.48

E 5.02 2.48 2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

6 0

3 years ago

PLS HELP ME I HAVE TILL 12

MariettaO [177]

Answer:

alright bud lets see hmm.... the answer is a. 90.5kpa

Explanation:

92.3 kPa - 1.82 kPa = 90.5 kPa

keep up your hope also corrected me if a am wrong in anyway! :)

3 0

3 years ago