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3 years ago

Arrange the statements made during a process to correspond with the scientific method.

Chemistry

1 answer:

mina [271]3 years ago

8 0

Answer:

The statements made during a process to correspond with the scientific method is arranged by sequence.

Explanation:

The accurate classification would be,

- You mention a sparkling silvery speck in a rock.

- The instrument presents proof that the specimen is zinc.

- You speculate the silvery speck could be gold.

- You practice an instrument to classify elements in the sparkling material.

- You presume that the material is zinc.

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If the ratio of the length of a rectangle to its width is 3 to 2, what is the

borishaifa [10]

Answer:

Length of a rectangle whose width is 4 inches is

6 inches

Explanation:

Ratio of length to its width is 3 to 2

Let length = L

Let Width = B

L: B = 3 : 2 (given)

$\frac{L}{B} =\frac{3}{2}$

Width ,B = 4

Insert B in above equation

$\frac{L}{B} =\frac{3}{2}$

$\frac{L}{4} =\frac{3}{2}$

cross multiply,

$L = \frac{3}{2}\times4$

L = 6 inches

Conversion(if required) ,

1 inch = 2.54 cm

6 inch = 2.54 (6)

L = 15.24 cm

8 0

3 years ago

11.23 mm = ____ m 1,123 m 0.01123 m 0.1123 m 11,230 m

kaheart [24]

For every meter, the equivalent measurements is 1000 millimeters. Hence in the problem where the number of millimeters is given, we divide the number by 1000 to get the number of meters. The answer here is 0.01123 m.

3 0

3 years ago

Read 2 more answers

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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3 years ago

What actions affect blood pH

galben [10]

Integestion, vomiting, diarrhea, lung function, endocrine function, kidney function

4 0

3 years ago

30 POINTS!!! NEED ASAL

nata0808 [166]

Answer:

ITS THE LAST ONE AND THE SECOND ONE

Explanation:

4 0

3 years ago