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2 years ago

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

Chemistry

1 answer:

Elodia [21]2 years ago

7 0

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

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How many moles of copper Il sulfide can be produced from 2.8 moles of sulfur? Cu + S --> CuS

Kamila [148]

Answer: 2.8 moles of copper (Il) sulfide $(CuS)$ will be produced from 2.8 moles of sulphur.

Explanation:

The balanced chemical reaction is:

$Cu+S\rightarrow CuS$

According to stoichiometry :

1 mole of $S$ produce = 1 mole of copper (Il) sulfide $(CuS)$

Thus 2.8 moles of $S$ will produce= $\frac{1}{1}\times 2.8=2.8moles$ of copper (Il) sulfide $(CuS)$

Thus 2.8 moles of copper (Il) sulfide $(CuS)$ will be produced from 2.8 moles of sulphur.

5 0

3 years ago

Which of the following compounds share electrons? NaCl CO CsF KBr

ArbitrLikvidat [17]

Covalent compounds share with each other, whilst Ionic compounds take from each other.

Covalent compounds are composed of 2 or more NONMETALS.

So, it would be CO, since the rest of the compounds contain metals in them.

7 0

3 years ago

A discus thrower throws a 1.6kg discus at 25m/s what's the kinetic energy?

Ilya [14]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

where

m is the mass

v is the velocity

From the question

m = 1.6 kg

v = 25 m/s

We have

$k = \frac{1}{2} \times 1.6 \times {25}^{2} \\ = 0.8 \times 625 \\ = 500$

We have the final answer as

Hope this helps you

6 0

3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago

When you push or pull something, you are creating

Katarina [22]

force but could be tension

3 0

2 years ago