2. Sketch an example where an object appears to be moving from one reference frame but stationary from another reference frame

Nick needs 3.5 moles of sodium fluoride (NaF) for a solution he is mixing in the lab. How many grams of NaF does Nick need to me

Minchanka [31]

Answer:

147 g

Explanation:

The number of moles of a substance is the ratio of the mass of the substance and the molar mass of that substance. Mathematically,

mole = mass of substance/molar mass

In this case, Nick needs 3.5 moles of NaF, the molar mass of NaF is calculated as:

23 + 19 = 42 g/mol (Note: Na = 23, F = 19)

Hence,

mass of 3.5 mole NaF = mole x molar mass

= 3.5 x 42 = 147 g.

Hence, Nick would need to measure out 147 g NaF using a suitable weighing balance.

5 0

3 years ago

Is the distilled water/acidic/basic/netural?

SVEN [57.7K]

Distilled water is neutral, because the H+ and OH- ions are balanced. Litmus is a natural acid-base indicator extracted from a type of lichen.

7 0

3 years ago

In the presence of an emulsifying agent, a mixture of oil and water becomes a _____.

Paladinen [302]

In the presence of an emulsifying agent, a mixture of oil and water becomes a colloidal dispersion.
Colloidal dispersion otherwise colloid is a system, in which discrete particles, droplets or bubbles of a dispersed phase (in this case oil), whose size at least in one dimension is in the range from 1 to 1000 nm are distributed in the other, usually continuous phase - dispersion medium (in this case water) differing from the dispersed phase in composition or state of aggregation.

8 0

3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago

What is the charge associated with each side of the HCl molecule?

Alexandra [31]

HCl is a polar molecule with the hydrogen part being partial positive while the chlorine end being partial negative. This is because hydrogen has an electronegativity of 2.1, and chlorine has an electronegativity of 3.0. This means that chlorine attracted most of the electron cloud of molecule hence is the negative dipole, The dipole moment of HCl is 1.08 D (debyes). A Debye is equal to 3.34 x 10-30 coulomb-meters (C-m). The charge of each molecule is o.176+ for H and 0.176- for the Cl

3 0

3 years ago

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