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brilliants [131]

4 years ago

5

two substances are poured into a cup and a reaction occurs. a thermometer is paced in the cup and the temperature in the surroun

ding area begins to drop. based on this information, was the reaction exothermic or endothermic?

1 answer:

STALIN [3.7K]4 years ago

3 0

Endothermic because it 'sucks' the energy and that's why temperature drops

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Thomson's model suggested that electrons were mixed into the sphere of an atom much like raisins ?

krek1111 [17]

True, because his model has the electrons in the sphere of the atom, Thomson’s model was called the “Plum Pudding” model because it was popular and relatable at his time.
Hope this helps!! ^v^

8 0

4 years ago

How many neutrons does the following nuclide have? 253/100 Fm A. 253 B. 100 C. 153 D. none of these

lara [203]

I believe the correct answer is C. The number of neutrons in the <span>253/100 Fm would be 153. The numbers beside Fm are used to determine this. Two hundred fifty three represents the mass number which is the sum of protons and neutrons in the nuclide while 100 represents the atomic number which is the number of protons. Therefore,

253 = neutrons + 100
neutrons = 153</span>

3 0

3 years ago

Read 2 more answers

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F- per 70 kg of body weight can cause death. Nevertheless, in orde

svetoff [14.1K]

Answer:

$\boxed{\text{(a) 14 L; (b) 711 kg}}$

Explanation:

(a) Litres of water

$\text{Mass of F}^{-} = 8.5 \times 10^{7}\text{ gal} \times \dfrac{\text{0.2 mg F}^{-}}{\text{1 kg}} = \textbf{14 mg F}^{-}\\\\\text{Volume of water} = \text{14 mg F}^{-} \times \dfrac{\text{1 L water}}{\text{1 mg F}^{-}} = \textbf{14 L water}\\\\\text{The person would have to consume $\boxed{\textbf{14 L}}$ of water per day}$

(b) Mass of NaF

$\text{Volume} =8.5 \times 10^{7} \text{gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} = 3.22 \times 10^{8}\text{ L}\\\\\text{Milligrams of F}^{-} = 3.22 \times 10^{8}\text{ L} \times \dfrac{\text{ 1 mg F}^{-}}{\text{1 L}} = 3.22 \times 10^{8}\text{ mg F}^{-}\\\\\text{kilograms of F}^{-} = 3.22 \times 10^{8}\text{ mg F}^{-} \times \dfrac{\text{1 mg F}^{-}}{10^{6}\text{kg F⁻}} = \text{322 kg F}^{-}$

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

$\text{Kilograms of NaF}= \text{322 kg F}^{-} \times \dfrac{\text{41.99 kg NaF}}{\text{19.00 kg F}^{-}} = \text{711 kg NaF}\\\\\text{It will take } \boxed{\textbf{711 kg}} \text{ of NaF to treat the reservoir}$

7 0

3 years ago

Read 2 more answers

1.02 mg of a compound that is known to absorb at 340 nm is dissolved and diluted to 5.00 mL in a volumetric flask. A 1.00 mL ali

Vinvika [58]

Answer:

$6.97 X 10^{-4}$ M

Explanation:

The concentration of the analyte in the 5-mL flask would be $6.97 X 10^{-4}$ M

This is a problem of simple dilution that can be solved using the dilution equation;

C1V1 = C2V2,

where C1 = initial concentration, V1 = initial volume, C2 = final concentration, and V2 = final volume.

<em>In this case, the initial concentration (C1) is not known, the initial volume (V1) is 1.00 mL, the final concentration is 6.97 x 10-5 M, and the final volume is 10.00 mL.</em>

Now, let us make the initial concentration the subject of the formula from the equation above;

C1 = C2V2/V1. Solve for C1 by substituting the other parameters.

C1 = 6.97 x 10-5 x 10/1 = $6.97 X 10^{-4}$ M

7 0

3 years ago

Match the items

Nostrana [21]

1. D
2. F
3. A
4. E
5. B
6. C

4 0

3 years ago