All categories

3 years ago

The Earth is always _______ as it revolves around the sun.

Chemistry

2 answers:

EastWind [94]3 years ago

5 0

Answer

Rotating

Explanation:

The Earth is always being pulled towards the Sun by gravity.

The Earth is not moving fast enough to "escape" the Sun's gravity and leave the solar system, but it is going too fast to be pulled into the Sun. Therefore, it keeps going around and around( roating)- orbiting the Sun.

svet-max [94.6K]3 years ago

5 0

Answer:

Rotating around the sun

You might be interested in

Which of the following is not true of acids?

NeTakaya

Answer is: 2. dillute acids feel slipper.

1) Acids are corrosive is correct. For example hydrochloric acid (HCl) will react with most metals.

2) Dillute acids feel slippery is not correct. Bases, for example solution of sodium hydroxide feels slipery.

3) Acids have a distinctly sour taste is correct. For example, vinegar is mixture of acetic acid (CH₃COOH) and water (H₂O). Vinegar is colourless liquid with sour taste and pungent smell, freezing point of the vinegar is lower than glacial acetic acid.

4) Acids have more hydronium ions than hydroxide ions is correct. Because acid gives a lot of hydrogen cations (H⁺), pH (pH = -log[H⁺]) is lower than seven (acidic solution).

5 0

3 years ago

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$