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3 years ago

How many hydrogen atoms are in water, H2O?

1 answer:

6 0

So in every water molecule there is H2 and O, which means there are 2 hydrogens and 1 oxygen in every water molecule. Therefor, the answer is 2.

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What demonstrates the flow of energy and materials through

Tcecarenko [31]

A. Food chain

A food web also shows the flow of energy and materials through an ecosystem but in a complex web, not a single chain. A biogeochemical cycle does not deal with the flow of energy/materials through an ecosystem at all.

3 0

1 year ago

A car is initially moving at 20 m/s east and a little while later it is moving at 10 m/s north. Which of the following best desc

Nina [5.8K]

Answer:d

Explanation:

Given

First car is moving towards east with velocity 20 m/s

$\vec{v_1}=20\hat{i}$

then it turns towards north then velocity is

$\vec{v_2}=10\hat{j}$

suppose car takes t sec to change its path so average acceleration is given by

$a=\frac{v_2-v_1}{t}$

$a=\frac{1}{t}(10\hat{j}-20\hat{i})$

So average acceleration is towards North of west.

5 0

3 years ago

Compressional forces within the crust can produce:

brilliants [131]

<span>tension, compression, and shearing and can i get brainliest plz</span>

4 0

3 years ago

Listed below are the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones. Use the given data to const

Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:

1. Median

2. Lower quartile

3. Upper quartile

4. Minimum value

5. Maximum value

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0

2 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago