All categories

3 years ago

a collection of nerve fibers through which impulses travel between the central nervous system and other parts of the body

1 answer:

4 0

A collection of nerve fibers through which impulses travel between the central nervous system and other parts of the body are the nerves.

You might be interested in

For the closed system below: Cu(s) + 2Ag+(aq) 2Ag(s) + Cu2+(aq) How would you know if the system is at equilibrium?

avanturin [10]

C. the mass of Cu(s) remains constant

7 0

3 years ago

Read 2 more answers

Select the correct mass for each of the samples described below.

NISA [10]

To determine the mass of each compound, we need to convert them into units of moles first. We use avogadros number to do this and use the molar mass of the compounds to convert to mass.

3.01 × 1023 molecules H2O ( 1 mole / 6.022x10^23 kg) (18.02 g / 1 mole) = 9.01 g

8.32 × 1020 formula units CaBr2 ( 1 mole / 6.022x10^23 kg) (199.89 g / 1 mole) = 0.28 g

1.93 × 1026 molecules XeF6 ( 1 mole / 6.022x10^23 kg) (245.28 g / 1 mole) = 78610.16 g

8 0

4 years ago

Read 2 more answers

What is atom?????????????

kiruha [24]

Answer:

An atom is a particle of matter that uniquely defines achemical element. An atom consists of a central nucleus that is usually surrounded by one or more electrons. ... The nucleus is positively charged, and contains one or more relatively heavy particles known as protons and neutrons. A proton is positively charged.

6 0

3 years ago

Read 2 more answers

How many grams of ethanol (density 0.80 g/cm 3) should be added to 5.0 g of chloroform (density 1.5 g/cm 3) if the resulting mix

kaheart [24]

Answer:

9.2 grams of ethanol should be added.

Explanation:

Let the mass of ethanol be x

Density of an ethanol = $d_1=0.80 g/cm^3$

Volume of the ethanol = $V_1=\frac{x}{0.80 g/cm^3}$

Mass of chloroform = 5.0 g

Density of an chloroform= $d_2=1.5 g/cm^3$

Volume of the ethanol = $V_2=\frac{5.0 g}{1.5 g/cm^3}=\frac{1}{3}$

Mass of the mixture,m = x + 5.0 g

Density of the mixture = $d=1.2 g/cm^3$

Volume of the mixture= $V=\frac{m}{1.2 g/cm^3}=\frac{x+5.0 g}{1.2 g/cm^3}$

$V=V_1+V_2$

$\frac{x+5.0 g}{1.2 g/cm^3}=V_1+\frac{1}{3}$

$\frac{x+5.0 g}{1.2 g/cm^3}=\frac{x}{0.80 g/cm^3}+\frac{1}{3}$

Solving for x , we get

x = 9.2 g

9.2 grams of ethanol should be added.

8 0

3 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

3 years ago