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klio [65]
3 years ago
14

Fish and reptiles share which of these traits?

Chemistry
1 answer:
Naddika [18.5K]3 years ago
8 0
I think it might be a or b but I’m 97% sure the answers is b
You might be interested in
How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain​
11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

<em>mg NaCl:</em>

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0
3 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
The density of benzene at 15 ∘C is 0.8787 g/mL. Calculate the mass of 0.1200 L of benzene at this temperature.
Pavel [41]
Use the density to convert volume into mass. 

since the density is in g/ml and the volume was given in Liters, we need to first convert the Liters into mililiters. just multiply by 1000 or move the decimal three times.

0.1200 Liters= 120.0 mL

120.0 mL (0.8787 grams/ 1 mL)= 105 grams
3 0
3 years ago
Read 2 more answers
2c+02=2CO2. The moles of co2 produced when 0.25 moles of O2 react is?​
Sophie [7]
<h3>Answer:</h3>

\displaystyle 0.5 \ mol \ CO_2

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Moles
  • Compounds

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C + O₂ → 2CO₂

[Given] 0.25 moles O₂

[Solve] moles CO₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol O₂ → 2 mol CO₂

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.25 \ moles \ O_2(\frac{2 \ mol \ CO_2}{1 \ mol \ O_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                        \displaystyle 0.5 \ mol \ CO_2
7 0
3 years ago
Pre-lab questions 1. a concentration gradient affects the direction that solutes diffuse. describe how molecules move with respe
VLD [36.1K]

Just like how heat moves from a region of higher temperature to a region of lower temperature, molecules also tend to move from a region of higher concentration to a region of lower concentration. This is called natural diffusion and is naturally happening to reach stability.

7 0
4 years ago
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