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4 years ago

Tissues in the human body such as cardiac tissue are made up of?

2 answers:

6 0

Answer : Tissues are usually made up of cells. Cardiac tissues or to be specific Cardiac muscle tissue are a kind of tissue which is made up of several interlocking cardiac muscle cells, or fibers, present in heart, which gives the tissue its properties.

Each cardiac muscle fiber is assumed to contain a single nucleus which is striated, or striped, as it appears to have light and dark bands when seen through a microscope.

Lynna [10]4 years ago

4 0

Cardiac muscle tissues are made up of cardiac muscle cells or muscle fibers that give the distinct characteristics and function of the heart.

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Which question would most likely be studied by a chemist?

lbvjy [14]

Answer:

Explanation:

fff

3 0

4 years ago

Copper has two naturally occurring isotopes, 63Cu (isotopic mass 62.9296 amu) and 65Cu (isotopic mass 64.9278 amu). If copper ha

Semenov [28]

Answer:

Percentage of first isotope = 69.152 %

Percentage of second isotope = 30.848 %

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

For first isotope:

Let % = x %

Mass = 62.9296 amu

For second isotope:

% = 100 - x % (Since, there are only two isotopes)

Mass = 64.9278 amu

Average mass = 63.546 amu

Thus,

$63.546=\frac {x}{100}\times {62.9296}+\frac {(100-x)}{100}\times {64.9278}$

Solving,

1.9982 x = 138.18

Thus,

<u>Percentage of first isotope = x = 69.152 %</u>

<u>Percentage of second isotope = 100 - x % = 30.848 %</u>

5 0

4 years ago

How many molecules of nitrogen gas are in 15 L of nitrogen gas

uranmaximum [27]

Answer:

$3.98\times 10^{23}$

Explanation:

Since we have $N_A$ number of molecules in 22.4 L, in 15 L we have:

$\frac{15}{22.4}\times 6.023 \times 10^{23}$ molecules.

5 0

3 years ago

In the laboratory, a quantity of I2 was reacted with excess H2 to give 1.26 moles of HI. It is also known that the percent yield

Anon25 [30]

Answer:

1.008moles of iodine

Explanation:

Hello,

This question requires us to calculate the theoretical yield of I₂ or number of moles that reacted.

Percent yield = (actual yield / estimated yield) × 100

Actual yield = 1.2moles

Estimated yield = ?

Percentage yield = 84%

84 / 100 = 1.2 / x

Cross multiply and solve for x

100x = 84 × 1.2

100x = 100.8

x = 100.8/100

x = 1.008moles

1.008 moles of I₂ reacted in excess of H₂ to give 1.2 moles of HI

5 0

4 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago