All categories

3 years ago

Please this is killing me I have no idea how to answer it it’s due in 10 mins please please help

Chemistry

1 answer:

klasskru [66]3 years ago

6 0

I think it’s 45g because

the maximum amount of sugar in a 100ml water is 180 and 25% of 180 is 45

You might be interested in

I need the answer please

padilas [110]

Answer:

transmission?

Explanation:

6 0

3 years ago

Suppose that you take a sample of 500 g of ocean water and let the water evaporate. The mass of the remaining salts is 17 g. Wha

yanalaym [24]

<u>Answer:</u> The salinity of ocean water is 34 ppt.

<u>Explanation:</u>

Salinity is defined as the number of grams of salts present per kilogram of seawater. It is expressed in parts per thousand.

The formula used to calculate the salinity of ocean water or sea water is:

$\text{ Parts per thousand}=\frac{\text{Mass of salts}}{\text{Mass of sea water}}\times 1000$

We are given:

Mass of salts = 17 g

Mass of sea water = 500 g

Putting values in above equation, we get:

$\text{ Parts per thousand}=\frac{17g}{500g}\times 1000\\\\\text{ Parts per thousand}=34$

Hence, the salinity of ocean water is 34 ppt.

8 0

3 years ago

If false rewrite the statements to make them correct True or False.

MaRussiya [10]

1. Only particles in the solid state are not in motion 2. The particles of a solid have less energy than gas 3. In a glass of iced tea the sugar in iced tea are the solvent and the water is the solute.

3 0

3 years ago

For the following reaction, 7.43 grams of methane (CH4) are allowed to react with 27.6 grams of carbon tetrachloride. methane (C

NikAS [45]

Answer:

30.4 grams of CH2Cl2 will be formed

Explanation:

Step 1: data given

Mass of methane = 7.43 grams

Molar mass of methane (CH4) = 16.04 g/mol

Mass of CCl4 = 27.6 grams

Molar mass of CCl4 = 153.82 g/mol

Step 2: The balanced equation

CH4 + CCl4 → 2CH2Cl2

Step 3: Calculate moles

Moles = mass / molar mass

Moles methane = mass methane / molar mass methane

Moles methane = 7.43 grams / 16.04 g/mol

Moles methane = 0.463 moles

Moles CCl4 = 27.6 grams / 153.82 g/mol

Moles CCl4 = 0.179 moles

Step 4: Calculate limiting reactant

For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

CCl4 is the limiting reactant. It will completely be consumed. (0.179 moles). Ch4 is in excess. There will react 0.179 moles. There will remain 0.463 - 0.179 = 0.284 moles

Step 5: Calculate moles CH2Cl2

For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

For 0.179 moles CCl4 we'll have 2*0.179 = 0.358 moles CH2Cl2

Step 6: Calculate mass CH2Cl2

Mass CH2Cl2 = 0.358 moles* 84.93 g/mol

Mass CH2Cl2 = 30.4 grams

30.4 grams of CH2Cl2 will be formed

3 0

4 years ago

The scientific method is sometimes seen as more of a cycle than a linear process. Which statement gives a major

irakobra [83]

Answer:

Explanation:

When a conclusions is shared it gives room for other scientist with other ideas or different perspective to raise up new questions

4 0

3 years ago