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3 years ago

Is an apple seed a compound or a mixture?

Chemistry

2 answers:

vaieri [72.5K]3 years ago

5 0

Answer:

Compound

Explanation:

compound: Substance that is made up of more than one type of atom bonded together

mixture: a combination of two or more elements of compound which are not chemically bonded

zheka24 [161]3 years ago

4 0

It’s a compound not a mixture.

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Identify the solution that contains both the solute and solvent in the solid phase. A) air B) NaCl C) bronze D) salt water

Iteru [2.4K]

Answer:prolly c

Explanation:

5 0

3 years ago

Choose two of the following scientists: Anton Lavoisier, John Dalton, JJ Thomson, Robert Millikan, Ernest Rutherford, James Chad

OverLord2011 [107]

Ernest Rutherford

J. J Thomson

Explanation:

<u>Ernest Rutherford</u>

In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.

Experiment

In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.

Discovery and reflection on the atomic theory

To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.

<u>J. J Thomson</u>

Experiment

In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.

Discovery and reflection on the atomic theory

From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:

they move in a straight line
they possess kinetic energy
they attract positive charges and repels negative charges

Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.

learn more:

Rutherford brainly.com/question/1859083

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6 0

3 years ago

Determine the number of equivalents if a 3.89N solution contains 0.76 L of solution

ValentinkaMS [17]

Answer:

Explanation:

oriented C-2, and (3) the minimizing of the number of ... (2) L. A. Mitscher, J. K. Paul, and L. Goldman,Experientia, 19, 195. (1963). ... SOzCeHiBr)3 in 147 ml. of anhydrous methanol containing 0.37 ... bicarbonate and saturated sodium chloride solution, and dried ... determined in 2% chloroform solution; infrared spectra on.

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3 years ago

A battery can provide a current of 4.60 A at 3.40 V for 2.50 hr. How much energy (in kJ) is produced? 1st attempt kJ Energy

Lunna [17]

Answer:

The energy produced equals 140.760 kJ

Explanation:

The relation between power, current and voltage is

$Power=Current\times Voltage$

Applying the given values in the relation above we get

$Power=4.60\times 3.40=15.64W$

Now Since $Power=\frac{Energy}{Time}\\\\Energy=Power\times Time$

Again applying the calculated values we get

$Energy=15.64\times 2.50\times 3600=140760Joules=140.76kJ$

4 0

4 years ago

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

4 years ago