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Gekata [30.6K]
2 years ago
12

Round 0.0011675cg to four significant figures

Chemistry
1 answer:
atroni [7]2 years ago
8 0

Answer:

answer is 0.001168 is correct

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You use a 15.0 gram piece of aluminum foil to cover a pan in the oven. The specific heat for aluminum is c = 0.900 J/g o C. If t
SVEN [57.7K]

Answer:

Best regards.

Explanation:

Hello,

In this case, we relate the heat, mass, heat capacity and temperature when a thermal change is carried out as shown below:

Q=mCp(T_{final}-T_{initial})

Now, for the given data, we compute the absorbed heat (due to the temperature increase) as follows:

Q=15.0g*0.900\frac{J}{g^oC}*(350^oC-25^oC) \\\\Q=4.39x10^3J=4.39kJ

Best regards.

8 0
2 years ago
Read 2 more answers
Graphing is an important procedure used by scientist to display the data that is collected during a controlled experiment. There
astraxan [27]

Bar graph or bar chart displays data graphically using bars

The correct values are;

1. The independent variable is the <u>month</u> of the year

2. The dependent variable is the <u>number of deer</u>

3. The title is:  <u>The number of deer counted per month</u>

The reason the above values are correct are as follows:

<u></u>

In a tabular form is given as follows;

\begin{array}{|c|cc|}Month&&\# \ of \ deer \\Sept&&38\\Oct&&32\\Nov&&26\\Dec&&20\\ Jan &&15\\Feb&&12\end{array}\right]

A. Please find attached the required bar graph (chart) created with a desktop application

1. In an investigation to assign the cause of an effect is the cause of an effect, the independent variable is the causal variable responsible for the observed effect, and it is not affected by the effect. It is the variable that is not measured

In the given data, the month of the year is not affected by the number of deer, and it is not counted, to see if there is a change, therefore, the independent variable is the <u>month</u> of the year

2. The dependent variable, is the effect variable that is measured, and it depends on the dependent variable, which is the month

Therefore, the dependent variable is the <u>number of deer</u>

<u></u>

3. An appropriate title is one that captures the information intended to be obtained from the Bar graph

An appropriate title for the bar graph is; <u>The number of deer counted per month</u>

Learn more about graphs here:

brainly.com/question/20465674

brainly.com/question/12730146

4 0
2 years ago
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic aci
Anton [14]

Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid</em>

<em />

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] <em>(1)</em>

<em></em>

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M  * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] <em>(2)</em>

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

<h3>15.2mL of the 0.10M sodium formate solution</h3>

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

<h3>4.8mL of the 0.10M formic acid solution</h3>
4 0
3 years ago
What is the density of ethane gas, C2H6 at STP
yawa3891 [41]

Answer:

The answer is that ethane gas has a density of 1.34 g/L at STP

Explanation:

Ethane weighs 0.0013562 gram per cubic centimeter or 1.3562 kilogram per cubic meter, i.e. density of ethane is equal to 1.3562 kg/m³; at 0°C (32°F or 273.15K) at standard atmospheric pressure.

7 0
2 years ago
How many bromine atoms are present in 39.4 g of CH2Br2
strojnjashka [21]
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2 

<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
7 0
3 years ago
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