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Yakvenalex [24]
3 years ago
11

Great amounts of atomic energy are released when a _______reaction occurs.

Chemistry
1 answer:
Mariulka [41]3 years ago
8 0

Great amounts of atomic energy are released when a _______reaction occurs.

Great amounts of atomic energy are released when a chemical reaction occurs. The process can be an exothermic reaction or endothermic reaction depending on the substances involved in the reaction.

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luda_lava [24]
Am so sorry I don’t know how to help you hope you find the answer
7 0
3 years ago
Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio
AlexFokin [52]

Answer:

The equilibrium concentration of CH₃OH is 0.28 M

Explanation:

For the reaction: CO (g) + 2H₂(g) ↔ CH₃OH(g)

The equilibrium constant (Keq) is given for the following expresion:

Keq= \frac{(CH3OH)}{(CO) x (H2)^{2}} =14.5

Where (CH3OH), (CO) and (H2) are the molar concentrations of each product or reactant.

We have:

(CH3OH)= ?

(CO)= 0.15 M

(H2)= 0.36 M

So, we only have to replace the concentrations in the equilibrium constant expression to obtain the missing concentration we need:

14.5= \frac{(CH_{3}OH) }{(0.15 M) x (0.36 M) ^{2} }

14.5 x (0.15 M) x (0.36)^{2} = (CH₃OH)

0.2818 M = (CH₃OH)

6 0
3 years ago
The Environmental Protection Agency has determined that safe drinking
dmitriy555 [2]

Answer:

b) \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

c) Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

Explanation:

We assume that part a is test the claim. And we can conduct the following hypothesis test:

Null hypothesis: \mu =7

Alternative hypothesis \mu \neq 7

The statistic is to check this hypothesi is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

We know the following info from the problem:

\bar X = 6.4 , s=0.5, n =30

Replacing we got:

t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573

And the p value would be:

p_v= 2*P(Z

Since the p value is very low compared to the significance assumed of 0.05 we have enough evidence to reject the null hypothesis that the true mean is equal to 7 moles/liter

Part b

The confidence interval is given by:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

The confidence interval for this case is given (6.21, 6.59)

So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter

Part c

Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.

6 0
3 years ago
Orygon
grigory [225]

if 105 grams burns completely

therefore

105 ×22.4/48=49

8 0
3 years ago
Calculate the total amount of energy required to change 10.0 g of water from 35.0 degrees Celsius to 110. degrees Celsius.
Makovka662 [10]

Answer:

The total amount of energy required is 25,515.2 J.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When a system absorbs (or gives up) a certain amount of heat, it can happen that:

  • experience a change in its temperature, which involves sensible heat,
  • undergoes a phase change at constant temperature, or latent heat.

To calculate the latent heat the formula is used:

Q = m. L

Where

  • Q: amount of heat
  • m: mass
  • L: latent heat

To calculate sensible heat the following formula is used:

Q = m. c. ΔT

where:

  • Q: amount of sensible heat  
  • m: body mass
  • c: specific heat of the substance
  • ΔT: temperature range

In this case, you have in the first place a heat to raise the temp of the water from 35.0 C to 100 C, where the specific heat value for water is  4.184 \frac{J}{g*C}:

q1 = m*c*(Tfinal-Tinitial)

q1 = 10.0 g *(4.184 \frac{J}{g*C})* (100 - 35.0 C) = 2719.6 J

Now you have the heat to vaporize the water, where the heat of vaporization is 2259.36 \frac{J}{g}:

q2 = m*(heat of vaporization)

q2 = 10.0 g*(2259.36 \frac{J}{g}) = 22593.6 J

Finally, you have the heat to raise temp of steam to 110 C, where the specific heat value for steam is  2.02 \frac{J}{g*C} :

q3 = m*c*(Tfinal-Tinitial)

q3 = 10.0 g*(2.02 \frac{J}{g*C})*(110-100 C) = 202 J

The total amount of energy can be calculated as:

Q= q1 + q2 + q3

Q= 2719.6 J + 22593.6 J + 202 J

Q=25,515.2 J

<u><em>The total amount of energy required is 25,515.2 J.</em></u>

5 0
3 years ago
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