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3 years ago

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A river current has a velocity of 10 km/h related to the shore and a boat moves in the same direction as a current of 10 km/h re

lated to the River how can the velocity of the boat related to the shore fee calculated

2 answers:

STatiana [176]3 years ago

8 0

Answer: Relative velocity of the boat with respect to shore is 20 km/h.

Explanation;

Relative velocity of the river with respect to shore,V = 10 km/h

Velocity of the shore = 0 km/h

Relative velocity of the river with respect to shore:

$V=V_r-V_s$

$10 km/h=V_r-0km/h$

Velocity of the river $V_r$ = 10 km/h

Velocity of the boat = $V_b$ = ?

Relative velocity of the boat with respect to river ,V'= 10 km/h

$V'=V_b-V_r$

$10 km/h=V_b-10 km/h$

$V_b=20 km/h$

Velocity of the boat $V_b$ = 20 km/h

Relative velocity of the boat with respect to shore, V":

$V"=V_b-V_s=20 km/h-0 km/h=20 km/h$

Relative velocity of the boat with respect to shore is 20 km/h.

Diano4ka-milaya [45]3 years ago

6 0

20 km/hr. Velocity is magnitude and direction. If two vectors are going in the same direction, you just add the magnitude.

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Read 2 more answers

How many atoms are found in 3.45g of CO2?

USPshnik [31]

<u>Answer:</u> The number of carbon and oxygen atoms in the given amount of carbon dioxide is $4.72\times 10^{22}$ and $9.44\times 10^{22}$ respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of carbon dioxide gas = 3.45 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of carbon dioxide gas}=\frac{3.45g}{44g/mol}=0.0784mol$

1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.

According to mole concept:

1 mole of a compound contains $6.022\time 10^{23}$ number of molecules

So, 0.0784 moles of carbon dioxide gas will contain $1\times 0.0784\times 6.022\times 10^{23}=4.72\times 10^{22}$ number of carbon atoms and $2\times 0.0784\times 6.022\times 10^{23}=9.44\times 10^{22}$ number of oxygen atoms

Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is $4.72\times 10^{22}$ and $9.44\times 10^{22}$ respectively

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3 years ago

If 2 moles of salt is dissolved to form 1 liter of solution, what is the molarity of the solution?

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Answer:

2M

Explanation:

Molarity refers to the molar concentration of a solution. It can be calculated by using the formula as follows:

Molarity (M) = number of moles (n) ÷ volume (V)

Based on the information provided in this question, 2 moles of salt is dissolved to form 1 liter of solution. This means that n = 2mol, V = 1L

Molarity = n/V

Molarity = 2/1

Molarity = 2M

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2 years ago

Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4

Katena32 [7]

<u>Answer:</u> The mass of rust that can be removed is 1.597 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = $6.00\times 10^2mL$ = 600 mL = 0.600 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol$

For the given chemical reaction:

$Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)$

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = $\frac{1}{6}\times 0.06=0.01mol$ of ferric oxide (rust)

To calculate the mass of rust for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

$0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g$

Hence, the mass of rust that can be removed is 1.597 grams

5 0

3 years ago