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AlladinOne [14]
2 years ago
12

True or False: The Ring of Fire is a section along the Atlantic Ocean where you have active volcanoes

Chemistry
2 answers:
Romashka [77]2 years ago
6 0
False because the Atlantic Ocean dose not have active volcanoes besides that’s a ocean so no
Scilla [17]2 years ago
6 0
False it’s the parific ocean that has active volcanoes
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List some general characteristic of producers and list and a few examples
kotykmax [81]

Answer:

pooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooop

Explanation: fart

8 0
3 years ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
3 years ago
Please Help me if you can!:) i appreciate anything.
enot [183]

Explanation:

1

Number of nucleon =

Molarmassofnucleon

Massofatom

=

1.6726×10

−24

g/nucleon

3.32×10

−23

g

=19.8=20(approximately)

It is given that element comprises of 2 atoms

Hence,number of nucleon = 2×20=40

2

You have 4.70 mol H2O

There are two H atoms in 1 molecule H2O.

Therefore, there must be 2*4.70 = 9.40 mols H in 4.70 mols H2O.

How many mols O in 4.70 mols H2O? That's 4.70 mols, of course.

Said another way, you have 2 mols H for every 1 mol H2O and 1 mol O for every 1 mol H2O.

So for 50 mols H2O you have 100 mols H and 50 mol O.

7 0
2 years ago
What is meant when it is said that stars have a life cycle?
vodomira [7]
C) Stars get born, get older, then eventually die.
6 0
3 years ago
Read 2 more answers
ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n
ArbitrLikvidat [17]

Answer:

\boxed{\text{3.3 mL}}

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}

2. Moles of NaOH  

\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}

3. Molar concentration of NaOH

c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

c_{1}V_{1} = c_{2}V_{2}

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1   mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}

(b) Calculate the volume  of dilute solution

\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}

\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}

4 0
3 years ago
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