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3 years ago

Most scientific discoveries are made when scientists work collaboratively

Chemistry

1 answer:

postnew [5]3 years ago

7 0

I believe it’s B. At least thats what makes sense for me.

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HELP ASAP PLEASAE!!!

N76 [4]

I believe the answer is A. However, I would double check the formula.

4 0

3 years ago

Help me plz, I need help on this.

kirza4 [7]

Omg i lost everything ugh
To do it again

1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol

2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol

3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol

4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol

5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g

6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g

7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g

8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g

I cant believe i had to do this all over

4 0

3 years ago

Prove how the Law Conservation of Mass and Enery apply to Cellular Respiration

snow_lady [41]

Answer:

mass and energy is reused when those things happen

7 0

3 years ago

4. How would you calculate the poH of a base when given the<br><br> concentration of [OH]-?

Novay_Z [31]

Answer:

Use pOH = -log₁₀ [OH-]

Explanation:

pOH can be calculated from the concentration of hydroxide ions using the formular below:

pOH = -log₁₀ [OH-]

The pOH is the negative logarithm of the hydroxide ion concentration.

3 0

3 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

4 years ago