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Setler [38]
2 years ago
11

The prefix centi- means

Chemistry
2 answers:
olya-2409 [2.1K]2 years ago
5 0

100. Centi means a hundred.

oksano4ka [1.4K]2 years ago
4 0
Centi means one hundred.
You might be interested in
What is the percent composition of hydrogen in NH4HCO3?
Readme [11.4K]
To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent. 
So we get: 
\frac{5}{79} * 100
So you get about: 
0.063 * 100
6.3

So, the percent composition of Hydrogen in NH4HCO3 is 6.3%

Hope this Helps! :D


7 0
3 years ago
Read 2 more answers
Leon uses a pressure gauge to measure the air pressure in one of his car tires. The gauge shows that the pressure is 220 kilopas
mamaluj [8]

Answer:

53.8 L

Explanation:

Ideal gas law

PV=nRT

must be for volume so we arrange to V=nRT/P

V= (4.8)(8.31)(297)/220

3 0
3 years ago
Read 2 more answers
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
In step 2, of the experiment, the procedure uses 3.0M NaOH. However, the student notices that the only solution of NaOH is conce
Luda [366]

Answer:

We need 78.9 mL of the 19.0 M NaOH solution

Explanation:

Step 1: Data given

Molarity of the original NaOH solution = 19.0 M

Molarity of the NaOH solution we want to prepare = 3.0 M

Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L

Step 2: Calculate volume of the 19.0 M NaOH solution needed

C1*V1 = C2*V2

⇒with C1 = the concentration of the original NaOH solution = 19.0 M

⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED

⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M

⇒with V2 = the volume  of the NaOH solution we want to prepare = 500 mL = 0.500 L

19.0 M * V2 = 3.0 M * 0.500 L

V2 = (3.0 M * 0.500L) / 19.0 M

V2 = 0.0789 L

We need 0.0789 L

This is 0.0789 * 10^3 mL = 78.9 mL

We need 78.9 mL of the 19.0 M NaOH solution

8 0
3 years ago
What is the power of 10 when 75,00 is written in scientific notation
blagie [28]

Answer:

7.5X1000

scientific notation

6 0
2 years ago
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