What is oxidized in a galvanic cell? O A. The salt bridge O B. The electrolytes O C. The cathode D. The anode

Which of the following is a strong acid?

4vir4ik [10]

H2SO4 is referred to as a strong acid and is denoted as option A.

This refers to any substance which tastes sour when in water and changes the color of blue litmus paper to red. It is usually very corrosive and are used in industries for different functions.

H2SO4 is referred to as a strong acid because it dissociates completely in its aqueous solution or water.

Read more about Acid here brainly.com/question/25148363

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3 0

2 years ago

Lion 7

Sidana [21]

Answer:

A.

Explanation:

You should NEVER eat or drink anything in a lab area. You never know what chemicals or gases are in the lab, and they can harm you.

Wearing a drawstring hoodie won't protect you from chemicals.

Don't wait to clean up chemicals, immediately get a teacher and clean it up (follow the teachers instructions). You never know what has spilled, and if it is harmful or not, or if there is a certain procedure to clean it up.

Don't change the equipment in the middle of an experiment. This can tamper with your results, and depending on what you are working with, this can be dangerous.

3 0

2 years ago

A sample of a compound containing boron (B) and hydrogen (H) contains 5.443 g of B and 1.522 g of H. The molar mass of the compo

Svetach [21]

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

first you divide the mass of BORON by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass.

So, 5.443/11 is about 0.5. Then, the RFM of H is 1, so the number of mole is 1.522/1=1.522.

Next, you get the number of moles in order is; 0.5 and 1.522. Now we need to look at the ratio between these numbers. 0.5 is smaller so we use it as the ratio of 1. next use 1.522/0.5 is 3.044 which has a greatest common factor of 3. so the empirical formula is BH3.

Now we are going to solve the molecular formula.

the molar mass ofthis compound is 30g, so we're going to find the RFM of the empirical formulsof BH3 first.

11+3=14.

now we see how many times 14 goes into 30. 30/14=2.14 which is about 2.

So now we need to times the subscript of the empirical formula by two.

thus, the molecular formula is B2H6.

To solve this kind of questions, there are many steps:Know what you are calculating about, it's about the molecular formula, so you need to find out the number of moles of each elements. then use the molar mass of the whole compound to calculate the molecular formula.

1) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4) then use the molarmass of the whole compound to find the molecular formula. molar mass of the whole compound/RFM(molar mass) of the empirical formula of elements= the number you need to multiply by the subscript of the empirical formula to get the molecular formula.

please tell me if i got anything wrong;)

4 0

2 years ago

What volume of hydrogen ( in L ) is produced from the complete reaction of 56.49 g of magnesium metal at STP

alexandr1967 [171]

Answer:

Mg = 24.30 g/mol) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Hint: 1 mole of gas at STP occupies 22.4 L

6 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago

What is oxidized in a galvanic cell? O A. The salt bridge O B. The electrolytes O C. The cathode D. The anode​

What is oxidized in a galvanic cell? O A. The salt bridge O B. The electrolytes O C. The cathode D. The anode