What else is produced when sodium carbonate decomposes? Na2CO3 → Na2O + ________

Find the linear function Cequalsf(F) that gives the reading on the Celsius temperature scale corresponding to a reading on the

padilas [110]

Answer:

C = (5/9) F - (160/9)

They both read equal at Z = - 40

Explanation:

We are looking for a linear function so we can write the following condition

Y = aX + b

Applying it to the exercise we got C = a F + b

Let's use the facts that C = 0 when F = 32 and C = 100 when F = 212

0 = 32 a + b (1)

100 = 212 a + b (2)

From (1) b = - 32 a , when we replace this in (2) we obtain a = (5/9)

and b = - (5/9)32 = - 160/9

Finally the linear function is C = (5/9) F - (160/9)

Both readings are equal at a Z number so

Z = (5/9) Z - 160/9

(4/9) Z = -160/9 and Z = - 40

6 0

3 years ago

In two or more complete sentences explain how to balance the chemical equation and classify its reaction type.

mihalych1998 [28]

Hey there!

C₅H₅ + Fe → Fe(C₅H₅)₂

Put a coefficient of 2 in front of C₅H₅ on the left side because there is a subscript of 2 after C₅H₅ in parenthesis on the right.

2C₅H₅ + Fe → Fe(C₅H₅)₂

Fe (iron) is already balanced since there is one on each side, so we don't need to change anything for that.

This is a synthesis reaction because two reactants, C₅H₅ and Fe, are yielding a single product, Fe(C₅H₅)₂.

Hope this helps!

6 0

2 years ago

Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres

____ [38]

Answer:

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature = $V_1=3.4 L$

Number of moles in the container = $n_1$

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = $V_2=5.11 L$

Number of moles in the container after addition= $n_2=2.28\times 10^{-2} mol$

Applying Avogadro's law:

$\frac{Volume}{Moles}=constant$ (at constant pressure and temperature)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}$

$n_1=1.51\times 10^{-2} mol$

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

8 0

3 years ago

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

Answer: The mass of original oxalic acid sample is 6.75 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0

2 years ago

Shoe companies imbed unique designs on the bottoms of the sneakers they manufacture. What is the main purpose of the designs on

Andreas93 [3]

It is to increase friction

3 0

3 years ago

What else is produced when sodium carbonate decomposes? Na2CO3 → Na2O + _________