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2 years ago

A hot cup of water has fewer molecules than a hot tub of water. This same cup of

2 answers:

6 0

Answer:eeded to change the temperature of one gram of a substance one degree ... iron(specific heat=0.12 cal/g C) and a cup of water both have the same temperature. ... If you drop a hot rock into a pail of water,the temperature of the rock and the water ... Therefore, water molecules have higher specific heat capacity than metals

Explanation:

Mnenie [13.5K]2 years ago

5 0

Answer:

Explanation:

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Why is this flower able to self-pollinate

Andrei [34K]

Answer:

A flower is self-pollinated if pollen is transferred to it from any flower of the same plant and cross-pollinated if the pollen comes from a flower on a different plant.

Explanation:

This is because each flower is capable of fertilizing itself by autogamy. Autogamy means that the male part of a flower sends pollen to the female part of the same flower.

3 0

3 years ago

1) A common experiment to determine the relative reactivity of metallic elements is to place a pure sample of one metal into an

Annette [7]

Answer: $Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4+Cu$

Explanation:-

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

As zinc is more reactive than copper, it could easily displace copper from its aqueous solution and thus leads to formation of zinc (II) sulfate and pure copper.

The chemical reaction can be represented as :

$Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4+Cu$

The phases are represented as (s) for solid sate, (l) for liquid state, (g) for gaseous state and (aq) for aqueous state.

7 0

3 years ago

Read 2 more answers

a sample of helium occupies a volume of 101.2 mL at a pressure of 790 mmHg. at what pressure would the volume be 120 mL?

AveGali [126]

Answer : The final pressure will be, 666.2 mmHg

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 790 mmHg

$P_2$ = final pressure = ?

$V_1$ = initial volume = 101.2 mL

$V_2$ = final volume = 120 mL

Now put all the given values in the above equation, we get:

$790mmHg\times 101.2mL=P_2\times 120mL$

$P_2=666.2mmHg$

Therefore, the final pressure will be, 666.2 mmHg

6 0

2 years ago

Many hospitals, and some doctors\' offices, use radioisotopes for diagnosis and treatment, or in palliative care (relief of symp

Phantasy [73]

Answer:

a) $^{131}_{53} I$

b) $^{192}_{77} Ir$

c) $^{153}_{62} Sm$

Explanation:

The symbols of the isotopes are written like

$^{A}_{Z} X$

where,

X is the element

A is the mass number (protons + neutrons)

Z is the atomic number (protons)

a) Iodine-131

The atomic number of iodine is 53. The mass number of this isotope is 131. The symbol is $^{131}_{53} I$ .

b) Iridium-192

The atomic number of iridium is 77. The mass number of this isotope is 192. The symbol is $^{192}_{77} Ir$ .

c) Samarium-153

The atomic number of samarium is 62. The mass number of this isotope is 153. The symbol is $^{153}_{62} Sm$ .

8 0

2 years ago

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,

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

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

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

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

$E_a=33813.28\ J/mol$

The activation energy for the decomposition = 33813.28 J/mol

8 0

2 years ago