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Xelga [282]
2 years ago
11

A(n) ______ Is a group of organs that work together to perform a com

Chemistry
1 answer:
soldier1979 [14.2K]2 years ago
5 0

Answer:

an organ system  Is a group of organs that work together to perform a com

Explanation:

that is the answer organ system

now plz give me 5 stars and press on the heart plz

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If an atom has 5 protons, how many electrons does it have?
IRISSAK [1]

Answer:

5

Explanation:

An atom has the same number of protons and electrons

8 0
3 years ago
Read 2 more answers
Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no rea
iragen [17]

Answer :

(A) The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

(B) The balanced molecular equation will be:

Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

(C) The balanced molecular equation will be:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

(D) The balanced molecular equation will be:

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}

Explanation :

Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.

Precipitation reaction : It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

Part A  : potassium carbonate and lead(II) nitrate

The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.

Part B : lithium sulfate and lead(II) acetate

The balanced molecular equation will be:

Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

In this reaction, lead sulfate is an insoluble salt and lithium acetate is a soluble solution.

Part C : copper(II) nitrate and sodium sulfide

The balanced molecular equation will be:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

In this reaction, Cuprous sulfide is an insoluble salt and sodium nitrate is a soluble solution.

Part D : strontium nitrate and potassium iodide

The balanced molecular equation will be:

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+SrI_2(aq)

In this reaction, strontium iodide and potassium nitrate are soluble solution.

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}

6 0
3 years ago
Both protons and neutrons (and their anti-particles) froze out:
VladimirAG [237]
I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge. 
6 0
3 years ago
What do the conclusions tell about the experiments?
Anni [7]

Answer:

the answer would be C of not C is B

4 0
2 years ago
Read 2 more answers
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
2 years ago
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