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Nezavi [6.7K]
3 years ago
10

Help !! Just number 10 ,11 ,12 thanks!

Chemistry
1 answer:
Marysya12 [62]3 years ago
6 0
10. the thing in the slide wasnt centered so the microscope magnefied on nothing 11. mive it tight 12. idk what a um is
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Which of the following statements is true about cellular respiration?
amid [387]
It turns sugar into H2O and CO2. So I guess you could say it breaks down food to release energy.
6 0
3 years ago
Read 2 more answers
1. Three
Bess [88]

Answer:

Explination:

Given Data:

                  Trail 1          Trial 2         Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm         446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e.  435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

5 0
3 years ago
Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3
krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction (\Delta S^o).

\Delta S^o=S_{product}-S_{reactant}

\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0{(NH_3)} = standard entropy of NH_3

\Delta S^0{(H_2)} = standard entropy of H_2

\Delta S^0{(N_2)} = standard entropy of N_2

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]

\Delta S^o=-198.3J/K

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0
3 years ago
Calculate the molarity of hydroxide ion in an aqueous solution that has a poh of 5.00
Simora [160]

Answer:

1.00 x 10^-5 M

Explanation:

pOH=-log[OH^-]

so

[OH^-]=10^(-pOH)

=10^-5

=1.00 x 10^-5 M

4 0
3 years ago
Which intermolecular force is characteristic of compounds which are liquids at room temperature and have relatively high boiling
SIZIF [17.4K]
Thew answer is hydrogen bonds. 
6 0
3 years ago
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