All categories

3 years ago

PLS HELP 20 POINTS

Chemistry

1 answer:

mixas84 [53]3 years ago

6 0

The answer is the third one because the animal died before the rock was there

You might be interested in

Find the number of moles of water that can be formed if you have 138 mol of hydrogen gas and 64 mol of oxygen gas.

charle [14.2K]

2H₂₍g₎ + O₂ ₍g₎→ 2H₂O

138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O

128 mol H₂O

6 0

4 years ago

Read 2 more answers

Use the References to access important values if needed for this question There are 12 eggs in a dozen. If a farmer's chickens p

bekas [8.4K]

Answer:

$3.4\times 10^3 eggs$ are produces are in a month.

Explanation:

Quantity of eggs produced by the chicken in a month = 284 dozens

1 dozen = 12 eggs

Number of eggs in a month:

$284 dozens = 284\times 12 eggs =3,408 eggs\approx 3.4\times 10^3 eegs$

$3.4\times 10^3 eggs$ are produces are in a month.

4 0

3 years ago

If 1.27 moles of bromine gas (Br2) react with excess phosphorus (P), how many moles of phosphorus tribromide (PBr3) will be prod

frez [133]

Answer:

0.85 mole of PBr3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Br2 + 2P —> 2PBr3

From the balanced equation above,

3 moles of Br2 reacted to produce 2 moles of PBr3.

Therefore, 1.27 moles of Br2 will react to produce = (1.27 x 2)/ 3 = 0.85 mole of PBr3.

Therefore, 0.85 mole of PBr3 is produced by the reaction.

8 0

4 years ago

Help!!! Acids and Bases

valentina_108 [34]

ACID: A substance that accepts a lone pair of electrons

BASE: A substance that donates a lone pair of electrons

:D

4 0

3 years ago

Read 2 more answers

What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of

Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 = -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m= molality

molality = $\frac{moles of solute}{mass of solvent in kilograms}$

putting the value in the equation:

molality= $\frac{1.31}{0.53}$

= 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

6 0

3 years ago