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Leviafan [203]
3 years ago
15

What are the advantages of using a computer over a graphing calculator

Chemistry
1 answer:
andrezito [222]3 years ago
5 0
The answer to this question is:

The advantages of using a computer over a graphing calculator is the the "Computer larger amount of data" then the graphing calculator 

Hope this helped, Laffertytyler20
Your Welcome :)
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Define density of a substance.
ozzi

Answer:

density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre. ... Density can also be expressed as kilograms per cubic metre (in metre-kilogram-second or SI units).

Explanation:

4 0
3 years ago
Read 2 more answers
Large mountain ranges rising up from the ocean floor are called
mote1985 [20]

Answer:

c mid ocean ridges

Explanation:

google def-A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary.

a basin is like a bowl and a plain is flat.

7 0
3 years ago
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1. (8pt) Using dimensional analysis convert 600.0 calories into kilojoules
Ivanshal [37]

Answer:

1. 2.510kJ  

2. Q = 1.5 kJ

Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

1cal=4.184J\\\\1kJ=1000J

Then, we perform the conversion as follows:

600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ

2. Here, we use the general heat equation:

Q=mC(T_2-T_1)

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Q=236g*0.24\frac{J}{g\°C} (34.9\°C-8.5\°C)\\\\Q=1495.3J*\frac{1kJ}{1000J} \\\\Q=1.5kJ

Regards!

7 0
3 years ago
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13
Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\

  =\frac{1571}{10}\\\\=157.1

Calculating the standardn:

\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\

Please find the correct equation in the attached file.

=28.195

For point a:

=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\

For point b:

=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M

For point c:

= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\  M

It is calculated by using the slope value that is 1.75 \times 10^9 M^{-1}. The slope value 1.75 \times 10^9 M^{-1}is ambiguous.

7 0
3 years ago
A 33.69 g sample of a substance is initially at 29.4 °c. after absorbing 1623 j of heat, the temperature of the substance is 110
Sliva [168]
Q= mcΔT
1623 = 33.69g x c x (110.8 - 29.4)
1623 = 2742.366 g•°C x c
c = 0.59j/g•°C
4 0
4 years ago
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