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What is the ratio of the lithium and the chlorine atoms in the compound? __________________

Alex

Johnny has 69420 sperm cells, he busts 420 sperm cells. How much sperm is left??

6 0

3 years ago

The equations are 2NaHCO3 -> Na2CO3 + CO2 + H20, NaHCO3 -> NaOH + CO2, 2NaHCO3 -> Na2O + 2CO2 + H20

cricket20 [7]

Answer:

The product made is Na2CO3.

The % yield is 91.8 %

Explanation:

Step 1: Data given

Mass of baking soda (NaHCO3) = 5.0 grams

Molar mass of NaHCO3 = 84.0 g/mol

Each of the equations calculated is left with 2.4g of NaOH, 1.86 g Na2O, and 3.18g Na2Co3.

Step 2: The balanced equations

2NaHCO3 → Na2CO3 + CO2 + H20

NaHCO3 → NaOH + CO2

2NaHCO3 → Na2O + 2CO2 + H20

Step 3: Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

Moles NaHCO3 = 5.0 grams / 84.0 g/mol

Moles NaHCO3 = 0.060 moles

Step 4: Calculate moles of products

For 2 moles NaHCO3 we'll have 1 mol Na2CO3

For 0.060 moles NaHCO3 we'lll have 0.060 / 2 = 0.030 moles Na2CO3

For 1 mol NaHCO3 we'll have 1 mol NaOH

For 0.060 moles NaHCO3 we'll have 0.060 moles NaOH

For 2 moles NaHCO3 we'll have 1 mol Na2O

For 0.060 moles NaHCO3 we'll have 0.030 moles Na2O

Step 5: Calculate mass of products

Mass = moles * molar mass

Mass of Na2CO3 = 0.030 moles * 105.99 g/mol = 3.18 grams

Mass of NaOH = 0.060 moles * 40.0 g/mol = 2.4 grams

Mass of Na2O = 0.030 moles *61.98 g/mol = 1.86 grams

Step 6: Calculate the percent yield

% yield = actual yield / theoretical yield

% yield Na2CO3 = (2.92 grams / 3.18 grams) *100% = 91.8 %

% yield NaOH = (2.92 grams / 2.4 grams ) *100% = 121.6 %

% yield of Na2O = (2.92 grams / 1.86 grams ) * 100% = 157 %

The product made is Na2CO3, the other reactions have a % yield greater than 100 %

7 0

3 years ago

Evaluate the exponential expression (−2)6.

Olegator [25]

A general exponential expression is something like:

A^n

This means that we need to multiply the number A by itself n times.

Using that we will get (-2)^6 = 64

With that definition, we can rewrite:

(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)

So we just need to solve the above expression.

Also, remember the rule of signs:

(-)*(-) = (+)

We will get:

(-2)*(-2)*(-2)*(-2)*(-2)*(-2) = [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]

= 4*4*4 = 16*4 = 64

Then we got:

(-2)^6 = 64

If you want to learn more, you can read:

brainly.com/question/17172630

6 0

3 years ago

Calculate the concentration of OH in a solution that contains 3.9 10-4 M H30 at 25°C. Identify the solution as acidic, basic or

Norma-Jean [14]

Answer:

Option A) 2.6 × 10⁻¹¹ M, acidic.

Explanation:

The ion-product constant of water $K_{\rm w}$ gives a relationship between $\rm [H_3O^{+}]$ and $\rm [OH^{-}]$ :

$K_{\rm w} = \rm [H_3O^{+}]\cdot [OH^{-}]$ .

The exact value of $K_{\rm w}$ depends on the temperature. Under $\rm 25^{\circ}C$ , $K_{\rm w} \approx 1^{-14}$ .

The question states that $\rm [H_3O^{+}] = 3.9\times 10^{-4}\; M$ . As a result,

$\begin{aligned} {\rm [OH^{-}]} &= \frac{K_{\rm w}}{[{\rm H_3O^{+}}]}\\ &\approx\frac{1^{-14}}{3.9\times 10^{-4}} && \rm {\leftarrow Only~under~25^{\circ}C.}\atop{}\\&\approx \rm 2.6\times 10^{-11}\; M\end{aligned}$ .

A solution is acidic if $\rm [H_3O^{+}] > [OH^{-}]$ .
A solution is neutral if $\rm [H_3O^{+}] = [OH^{-}]$ .
A solution is acidic if $\rm [H_3O^{+}] < [OH^{-}]$ .

$\rm [H_3O^{+}] > [OH^{-}]$ for this solution. As a result, this solution is acidic.

8 0

3 years ago

What happens when you put an ice-pop outside? This is for the people who don’t know

Aleonysh [2.5K]

Answer:

It melts

Explanation:

7 0

3 years ago

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