All categories

3 years ago

10 points !! Which can happen during a physical change?

1 answer:

5 0

I think that the answer is A because if you think about water for an example. When water turns into ice, it's still technically water, just the molecules are frozen. The water is physically changed. It could also see why you think it is B but I don't think so completely. It doesn't really make sense to me. So I'd choose A. It's definitely not C or D.

You might be interested in

What types of orbital overlap occur in cumulene? check all that apply. check all that apply. p/p overlap sp/sp2 overlap s/sp ove

egoroff_w [7]

Cumulene is an alkene hydrocarbon that has two or three double bonds. It is also known as allene. The simplest cumulene is butatriene. It is a rigid hydrocarbon due to the presence of double bonds.

Answer: The orbital that overlap are p/p overlap, s/sp2 overlap, sp/sp2 overlap and sp/sp orbital.

8 0

3 years ago

Read 2 more answers

Suppose 10 atoms of carbon (C) react with 20 atoms of oxygen (O). According to the Law of Conservation of Matter, how many atoms

vova2212 [387]

Answer:

D 30

Explanation:

whenever there is a reaction, the amount of atoms don't change, because matter can't be created or destroyed. Add the carbon and oxygen atoms together

3 0

4 years ago

Read 2 more answers

During transcription, RNA polymerase encounters the sequence of DNA

SVEN [57.7K]

Answer:

it 123456789 •-• ™ I h--e yu hehehe

3 0

3 years ago

Read 2 more answers

Determine the empirical formulas for compounds with the following percent compositions:

Elis [28]

Answer:

For a: The empirical formula of the compound is $P_2O_5$

For b: The empirical formula of the compound is $KH_2PO_4$

Explanation:

For a:

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

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

Step 1: Converting the given masses into moles.

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles$

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 1.406 moles.

For Phosphorus = $\frac{1.406}{1.406}=1$

For Oxygen = $\frac{3.525}{1.406}=2.5$

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = $1\times 2=2$

For Oxygen = $2.5\times 2=5$

Step 3: Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is $P_2O_5$

For b:

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

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

Step 1: Converting the given masses into moles.

Moles of Potassium = $\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles$

Moles of Phosphorus = $\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles$

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 0.735 moles.

For Potassium = $\frac{0.736}{0.735}=1$

For Hydrogen = $\frac{1.5}{0.735}=2.04\approx 2$

For Phosphorus = $\frac{0.735}{0.735}=1$

For Oxygen = $\frac{2.9375}{0.735}=3.99\approx 4$

Step 3: Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is $KH_2PO_4$

3 0

3 years ago

How many grams of CaCl, are needed to prepare 120 mL of 450 M Cach solution?

harkovskaia [24]

Answer:

gggyyhhhhjjjjjjhhhhhhhhhhhhhjjjjj

5 0

3 years ago