Ask question

Ask question

All categories

3 years ago

8

Many organ systems serve to help maintain a healthy body. The excretory system is an organ system which is composed of the sweat

glands, the lungs, the rectum, and the kidneys, which all function to remove waste material.

1 answer:

geniusboy [140]3 years ago

8 0

Answer:

Its the circulatory system

You might be interested in

Balance each of these equations.

Oksanka [162]

Answer:

A. 2NO + O2 -> 2NO2

B. 4Co + 3O2 -> 2Co2O3

C. 2Al + 3Cl2 -> Al2Cl6

D. 2C2H6 + 7O2 -> 4CO2 + 6H2O

E. TiCl4 + 4Na -> Ti + 4NaCl

8 0

2 years ago

Chlorine + sulfur dioxide + water = hydrochloric acid + sulfuric acid

scZoUnD [109]

The given example is a chemical reaction.

The contents (separated as reactants and products) :

$\begin{tabular}{c | l}Reactants & Products \\\cline{1-2}Chlorine & Sulfuric Acid \\Water & Hydrochloric Acid \\Sulfur Dioxide & \\\end{tabular}$

The written reaction is :

$\boxed {Cl + SO_{2} + H_{2}O \implies HCl + H_{2}SO{4}}$

<em>I hope it helped you solve the problem.</em>

<em>Good luck on your studies!</em>

7 0

1 year ago

A closed flask contains gas at STP. If the temperature is raised to 400 K what will the pressure be in atm?

balandron [24]

Answer:

The answer is 1.5 atm

Explanation:

5 0

2 years ago

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

NemiM [27]

Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

$Q_1=-Q_2$

Mass of steel= $m_1$

Specific heat capacity of steel = $c_1=0.452 J/g^oC$

Initial temperature of the steel = $T_1=2.00^oC$

Final temperature of the steel = $T_2=T=21.50^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2= 105 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=22.00^oC$

Final temperature of water = $T_2=T=21.50^oC$

$Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))$

On substituting all values:

$(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}$

<h3>25.0 grams is the mass of the steel bar.</h3>

3 0

3 years ago

Read 2 more answers

How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2

Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>

4 0

2 years ago