NAME THESE COUMPANDS

qwelly [4]

Answer:

Moving objects contain energy. The faster the object moves, the more energy it has. ... Energy can be converted from one form to another form. Kinetic energy can be distinguished from the various forms of potential energy.

Explanation:

8 0

2 years ago

2c8h18(g 25o2(g→16co2(g 18h2o(g how many moles of water are produced in this reaction

AleksandrR [38]

18 moles of water are produced in the above reaction.
Hope this helps you!

3 0

2 years ago

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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

3 years ago

A gymnast jumps from a great height onto a sealed mat filled with air.

ololo11 [35]

The gas particles squeeze closer together

6 0

2 years ago

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How many bromine atoms are present in 37.9 g of CH2Br2?

VARVARA [1.3K]

Answer:

The answer to your question is: 6.55 x 10 ²³ atoms of Br

Explanation:

CH2Br2 = 37.9 g

MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g

174 g of CH2Br2 ------------------ 160 g of Br2

37.9 g of CH2Br2 --------------- x

x = 37.9 x 160/174 = 34.85 g of Br

1 mol of Br ----------------- 160 g Br2

x ---------------- 174 g Be2

x = 174 x 1 /160 = 1.088 mol of Br2

1 mol of Br ----------------- 6.023 x 10 ²³ atoms

1.088 mol of Br ------------- x

x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms

4 0

2 years ago