All categories

3 years ago

Help please! Its due in a few minutes!!

Chemistry

2 answers:

Ahat [919]3 years ago

5 0

Is time up because it won’t let me do it

lesya [120]3 years ago

4 0

Oh okkkkkkkkkkkkkkkk

You might be interested in

A buffer is prepared by mixing 50.3 mL of 0.183 M NaOH with 128.8 mL of 0.231 M acetic acid. What is the pH of this buffer

kvv77 [185]

Answer:

A buffer system can be made by mixing a soluble compound that contains the conjugate ... 10.0 grams of sodium acetate in 200.0 mL of 1.00 M acetic acid.

Explanation:

7 0

3 years ago

Which three words best describe the composition of the inner planets? solid, smooth, giant rocky, solid, dense giant, dense, gas

Lapatulllka [165]

Answer:

solid, rocky, dense

Explanation:

3 0

4 years ago

Read 2 more answers

A biochemist conducted an experiment to follow the movement of glucose molecules in animal cells. Glucose with radioactively lab

Airida [17]

Answer:

Here is a link I made for you

http://chs.helenaschools.org/wp-content/uploads/sites/32/2018/05/2016-2017-AP-Biology-Exam-Review.docx

Explanation:

This explanes all of what you need along with helpful videos an other things I found intersting for you HOPE IT HELPS YOU YOUR WELCOME.

6 0

3 years ago

A particular brand of gasoline has a density of 0.737 g/mL at 25 ∘C. How many grams of this gasoline would fill a 13.6 gal tank

12345 [234]

Answer:

Mass, m = 37900 grams

Explanation:

Given that,

The density of a brand of gasoline is 0.737 g/mL at 25°C

We need to find mass of gasoline would fill a 13.6 gal tank.

Density is defined as mass per unit volume.

We know that, 1US gal=3.78L

So,

1US gal = 3785.41 mL

13.6 US gal = 51481.6 mL

So,

$d=\dfrac{m}{V}\\\\m=d\times V\\\\m=0.737\ g/mL\times 51481.6\ mL\\\\m=37941.93\ g$

m = 37900 g (in three significant figures)

So, the mass of the gasoline is 37900 grams.

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

4 years ago