Ask question

Ask question

All categories

3 years ago

6

A duck migrates South for the winter about 3000 miles each year. It takes them about 65 hours to travel this distance. What is t

he velocity of the duck?

1 answer:

olchik [2.2K]3 years ago

7 0

Answer:

46.15

Explanation:

You might be interested in

The neutrons of an atom have ______ charges and are located ____ the nucleus

djyliett [7]

Neutral charges and are found in the center? I'm not sure for the last answer

6 0

3 years ago

Read 2 more answers

Which particles may be gained lost or shared by an atom when it forms a chemical bond?

Maru [420]

Answer:

B.

Explanation:

electrons can be lost by one particle, and gained by another particle

6 0

3 years ago

When the equation Ca(OH)2 + HBr → (products) is completed and balanced, one term in the equation will be

NISA [10]

<span>Answer for the given question is CaBr2. Although, the given equation requires balancing of by adding one more HBr and one more H2o in resultant. The given equation will product at least one CaBr2. Hence the answer for the given equation is Calcium bromide i.e. CaBr2.</span>

4 0

3 years ago

What is the volume of 60 g of ether if the density of ether is 70 g/mL

NNADVOKAT [17]

V = 60.0 g/ 0.70 g/mL = 85.7 mL Hope this helps! ;D

4 0

3 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

3 0

3 years ago