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3 years ago

Emily is a competitive swimmer. Why does wearing a swim cap help Emily swim faster?

1 answer:

8 0

Well the problem is that Emily is capping. PERIODT jpjp
the answer is
B) the cap keeps the hair out of her eyes

I SWEAR IF IM WRONG THAN EMILY CAPPING CUZ MY CAP DECTOR IS GOING OFF

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An unknown metal absorbs 61 J of heat, and its temperature increases by 29∘C. What is the heat capacity of the metal?

aniked [119]

Answer:

2.103 J/C

Explanation:

Quantity of heat = Heat Capacity * Temperature change

Heat Capacity = Quantity of heat / Temperature Change

Heat Capacity = 61/29

Heat Capacity = 2.103 J/C

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3 years ago

You are sitting outside on a warm day with a cold glass of water. You notice water droplets starting to collect on the outside o

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Answer:

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Drugs they have been devoloped to kill bacteria are called

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Answer:

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Antibiotic

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How many moles of Au are in 312 g of Au?

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Answer:

1.583 moles

Explanation:

Rounded Atomic Mass of Au = 197 grams

$\frac{312}{197} =1.5837, 1.584$

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3 years ago

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⁻.

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

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4 years ago

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