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3 years ago

Name a food that contain an acid

Chemistry

2 answers:

Triss [41]3 years ago

5 0

Answer:

name a food that contain an acid

1. nuts

2. dairy

3. grains

3. meat

Vaselesa [24]3 years ago

5 0

 meat. 
grains. 
dairy. 
unsprouted beans. 
sunflower and pumpkin seeds. 
nuts. 
carbonated drinks. 
alcohol

these are the foods which contains acids.

hope its helps you 

have a great day keep smiling be happy stay safe.

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Give all possible ml values for orbitals that have each of the following: (a) l = 3; (b) n = 2; (c) n = 6, l = 1.

larisa86 [58]

Answer : All possible values of 'ml' for the following orbitals are:

(a) At l = 3, $m_l=+3,+2,+1,0,-1,-2,-3$

(b) l = 0, 1

At l = 0, $m_l=0$

At l = 1, $m_l=+1,0,-1$

(c) At l = 1, $m_l=+1,0,-1$

Explanation:

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from $(-l\text{ to }+l)$ . When l = 2, the value of $m_l$ will be -2, -1, 0, +1, +2.

Spin Quantum number : It describes the direction of electron spin. This is represented as $m_s$ The value of this is $+\frac{1}{2}$ for upward spin and $-\frac{1}{2}$ for downward spin.

(a) l = 3 then the value of 'ml' is,

At l = 3, $m_l=+3,+2,+1,0,-1,-2,-3$

(b) n = 2 then the value of 'ml' is,

l = 0, 1

At l = 0, $m_l=0$

At l = 1, $m_l=+1,0,-1$

(c) n = 6 and l = 1 then the value of 'ml' is,

n = 6

l = 0, 1, 2, 3, 4, 5

At l = 1, $m_l=+1,0,-1$

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4 years ago

If nitrogen at 130 kPa is allowed to expand at constant temperature until its pressure is 145.5 kPa, how much larger will the vo

LenaWriter [7]

Answer:

0.89 times

Explanation:

Let the initial volume of the tank be V.

Finally, we shall determine how much larger the final volume of the gas will be. This is illustrated below:

Initial pressure (P₁) = 130 kPa

Initial volume (V₁) = V

Final pressure (P₂) = 145.5 kPa

Final volume (V₂) =?

P₁V₁ = P₂V₂

130 × V = 145.5 × V₂

130V = 145.5 × V₂

Divide both side by 145.5

V₂ = 130V / 145.5

V₂ = 0.89V

Therefore, the final volume of gas will be 0.89 times the initial volume.

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3 years ago

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3 years ago

1. If there are 100 navy beans, 27 pinto beans and 173 blackeyed peas in a container, what is the percent abundance in the conta

kompoz [17]

The answers to the two questions are:

1. The percent abundance in the container which has 100 navy, 27 pinto, and 173 black-eyed peas beans is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The weighted average score for the scores of 85, 75, 96 obtained from the evaluations of exams (20%), labs (75%), and homework (96%) is 84.1.

1. The percent abundance by type of bean is given by:

$\% = \frac{n}{n_{t}} \times 100$ (1)

Where:

n: is the number of each type of beans

$n_{t}$ : is the total number of beans

The total number of beans can be calculated by adding the number of all the types of beans:

$n_{t} = n_{n} + n_{p} + n_{b}$ (2)

Where:

$n_{n}$ : is the number of navy beans = 100

$n_{p}$ : is the number of pinto beans = 27

$n_{b}$ : is the number of black-eyed peas beans = 173

Hence, the total number of beans is (eq 2):

$n_{t} = 100 + 27 + 173 = 300$

Now, the percent abundance by type of bean is (eq 1):

Navy beans

$\%_{n} = \frac{100}{300} \times 100 = 33.3 \%$

Pinto beans

$\%_{p} = \frac{27}{300} \times 100 = 9.0 \%$

Black-eyed peas beans

$\%_{b} = \frac{173}{300} \times 100 = 57.7 \%$

Hence, the percent abundance by type of bean is 33.3%, 9.0%, and 57.7% for navy bean, pinto bean, and black-eyed peas beans, respectively.

2. The average score (S) can be calculated as follows:

$S = e*\%_{e} + l*\%_{l} + h*\%_{h}$ (3)

Where:

e: is the score for exams = 85

l: is the score for lab reports = 75

h: is the score for homework = 96

$%_{e}$ $\%_{e}$ : is the percent for exams = 70.0%

$\%_{l}$ : is the percent for lab reports = 20.0%

$\%_{h}$ : is the percent for homework = 10.0%

Then, the average score is:

$S = 85*0.70 + 75*0.20 + 96*0.10 = 84.1$

We can see that if the score for each evaluation is 100, after multiplying every evaluation for its respective percent, the final average score would be 100.

Therefore, the weighted average score will be 84.1.

Find more about percents here:

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I hope it helps you!

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3 years ago

Name a food that contain an acid​

Name a food that contain an acid