What should the person do?

A piece of neutral litmus paper turned red in some grapefruit juice.  What does this show about the grapefruit juice

Elodia [21]

Answer:

grapefruit juice is an acid

Explanation:

A piece of neutral litmus paper turned red in some grapefruit juice. What does this show about the grapefruit juice

with litmus paper remember:

E R

S E

ACID

BLUE

acids turn litmus red, bases turn it blue

all citrus fruits :llemons, lime, grapefruit, oranges, etc are called

CITRUS FRUITS because they contain citric acid

7 0

1 year ago

20.00 g of aluminum (al) reacts with 78.78 grams of molecular chlorine (cl2) all of each reactant is completely consumed and a s

marysya [2.9K]

The mass of the product is 98.78 g.

The word equation is

aluminum + chlorine → product

20.00 g + 98.78 g → x g

If each reactant is completely consumed, the Law of conservation of Mass tells us the mass of the product must be 98.78 g.

8 0

2 years ago

Name this compound Cr2(CO3)3

kiruha [24]

Answer:

Molecular Weight

Explanation:

Chromium(III) Carbonate Cr2(CO3)3 Molecular Weight -- EndMemo.

5 0

2 years ago

The exoskeleton of a housefly is made up of

OverLord2011 [107]

Answer:

Chitin

Explanation:

A fly's main body is made of a thorax and an abdomen. These parts are covered by a hard exoskeleton made of a bony substance called chitin.

5 0

2 years ago

Read 2 more answers

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of

Studentka2010 [4]

Answer :

(A) The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

Part A:

The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

Part B:

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

Part C:

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0

2 years ago