All categories

2 years ago

5. Students hypothesize that temperature is a limiting factor for the germination of seeds. They carry out an experiment to

Chemistry

1 answer:

Lelechka [254]2 years ago

8 0

Answer:

keep adding to on each term its quit simple

You might be interested in

A valid lewis structure of __________ cannot be drawn without violating the octet rule. a valid lewis structure of __________ ca

Oxana [17]

A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.

Answer: IF3 (Iodine Trifluoride)

This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.

5 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

Best regards.

8 0

3 years ago

Please help me ASAP

MrRissso [65]

Answer:

<h2>answer will be D</h2>

change of state only 

8 0

3 years ago

What is the frequency of a photon that has a wavelength of 754 um?

Bogdan [553]

Answer:

0.39760273

Explanation:

I typed into calculator hope it's right.

6 0

2 years ago

43. A stock glucose standard has a concentration of 1,000 mg/dL. A 1/5 dilution of this standard is made. What would be the fina

Nat2105 [25]

The final concentration of the diluted standard is 0.2 mg/dL.

<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>

Using the dilution formula:

C1V1 = C2V2

where

C1 is initial concentration
V1 initial volume
C2 is final concentration
V2 is final volume.

Assuming a final volume of 100 mL, and since a 1/5 dilution is made:

C1 = 1.00 mg/dL

V1 = 20

C2 = ?

V2 = 100 mL

C2 = C1V1/V2

C2 = 20 × 1/100

C2 = 0.2 mg/dL

Therefore, the final concentration of the diluted standard is 0.2 mg/dL.

Learn more about dilution at: brainly.com/question/24881505

6 0

2 years ago