Fill in the blanka gas-like state of matter consisting of free electrons and atomic nuclei is _________

Write the ionic and half equation for H20.

Pavlova-9 [17]

2[H⁺] + [O²⁻] → H₂O

Explanation:

Half reactions are usually composed of components of oxidation-reduction (redox) reaction where one element is losing electrons while another is gaining so they can come together into a compound. The half-reaction shows part of a reaction where the combining elements are charged and the product, from the redox, is neutral.

In this case, the H⁺ is carrying the positive charge because it is donating an electron to O²⁻, which is the reason the O²⁻ is carrying the negative charge. Note that we need 2 H⁺ for every one O²⁻. The O is being reduced while the H is being oxidized.

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6 0

3 years ago

How hot, in degrees Celsius, would the air inside the balloon have to get in order for the balloon to lift off the ground? Assum

SVETLANKA909090 [29]

Answer:

The temperature in degress Celsius is 52.25°C

Explanation:

According the equation:

$V(\rho _{a} -\rho _{t})g=mg\\\rho _{a} -\rho _{t}=\frac{n}{V} =\frac{340}{2950} =0.115kg/m^{3}$

$\rho _{t}=\rho _{a} -0.115=1.2-0.115=1.085kg/m^{3}$

The temperature is:

$T=\frac{P}{\rho _{t} R} =\frac{1.013x10^{5} }{1.085*287.05} =325.25K=52.25$ °C

3 0

3 years ago

If a combination reaction takes place between rubidium and bromine, the chemical formula for the product is

lesya [120]

The chemical formula for the product is RbBr.

7 0

2 years ago

Consider the following reaction where Kc = 1.80×10-2 at 698 K:2 HI (g) H2 (g) + I2 (g)A reaction mixture was found to contain 0.

Iteru [2.4K]

Answer:

Answer for the given statements: (1) T , (2) F , (3) T , (4) F , (5) F

Explanation:

At the given interval, concentration of HI = $\frac{0.304}{1}M=0.304M$

Concentration of $H_{2}$ = $\frac{5.07\times 10^{-2}}{1}M=5.07\times 10^{-2}M$

Concentration of $I_{2}$ = $\frac{4.57\times 10^{-2}}{1}M=4.57\times 10^{-2}M$

Reaction quotient, $Q_{c}$ , for this reaction = $\frac{[H_{2}][I_{2}]}{[HI]^{2}}$

species inside third bracket represents concentrations at the given interval.

So, $Q_{c}=\frac{(5.07\times 10^{-2})\times (4.57\times 10^{-2})}{(0.304)^{2}}=2.51\times 10^{-2}$

So, the reaction is not at equilibrium.

As $Q_{c}> K_{c}$ therefore reaction must run in reverse direction to reduce $Q_{c}$ and make it equal to $K_{c}$ . That means HI(g) must be produced and $H_{2}$ must be consumed.

3 0

2 years ago

The CO2 produced in one round of the citric acid cycle does not originate in the acetyl carbons that entered that round. If the

aleksley [76]

Answer:

It will require<u> second round</u> of the cycle to release $14C0_2$

Explanation:

<u>Reason behind the requirement of second round of the cycle to release </u> $CO_2$ -:

The C4 carbon of succinyl CoA is acetyl from acetyl CoA. Succinyl CoA is converted to succinate, which is then converted to fumarate, fumarate, malate, and eventually oxaloacetate. 14C will be found in oxaloacetate at either C1 or C4. During the second round of the loop, each of these carbons will be converted to carbon dioxide.

7 0

2 years ago

Fill in the blanka gas-like state of matter consisting of free electrons and atomic nuclei is __________