When 412.5 g of calcium carbonate react with 521.9 g of aluminum fluoride, how many

Potassium (K) and calcium fluoride (CaF2) can both be classified as<br> (Please help)

nasty-shy [4]

Potassium and calcium fluoride are both metals

8 0

4 years ago

For the skeletal chemical equation BF3(g) + NaH(s) → B2H6(g) + NaF(s) what is the coefficient of B2H6 in the balanced equation?

KonstantinChe [14]

Answer:

Coefficient in front of the $B_2H_6$ in the balanced equation - 1

Explanation:

The unbalanced Chemical equation is shown below as:-

$BF_3+NaH\rightarrow B_2H_6+NaF$

On the left hand side,

There are 1 boron atom and 3 fluorine atoms and 1 sodium and hydrogen atoms.

On the right hand side,

There are 2 boron atoms and 6 hydrogen atoms and 1 sodium and fluorine atoms.

Thus,

leftside, $BF_3$ must be multiplied by 2 to balance boron and right side, $NaF$ must be multiplied by 6 to balance fluorine. Left side, $NaH$ must be multiplied by 6 to balance sodium and hydrogen atoms.

Thus, the balanced reaction is:-

$2BF_3+6NaH\rightarrow B_2H_6+6NaF$

<u>Coefficient in front of the $B_2H_6$ in the balanced equation - 1</u>

6 0

3 years ago

Read 2 more answers

What are

Vesna [10]

Answer:

carbohydrates

Explanation:

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7 0

3 years ago

Read 2 more answers

How many moles are in 2.5L of 1.75 M Na2CO3

mixas84 [53]

Answer: There are 4.375 moles in 2.5 L of 1.75 M $Na_2CO_3$

Explanation:

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

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

Molarity of solution = 1.75 M

Volume of solution = 2.5 L

Putting values in equation , we get:

$1.75M=\frac{\text{Moles of} Na_2CO_3}{2.5L}\\\\\text{Moles of }Na_2CO_3=1.75mol/L\times 2.5L=4.375mol$

6 0

3 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = $T_1= 21.36^oC$

Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

$\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol$

3 0

3 years ago