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Olenka [21]
2 years ago
6

How many oxygen atoms are in 3.70 g of quartz?

Chemistry
1 answer:
Ad libitum [116K]2 years ago
3 0

Answer:

The formula of Quartz is SiO2. So, 1 mole of Quartz will have 2 moles of oxygen atoms, i.e. 2 x 6.022 x 10^23 atoms of oxygen. The molar mass of quartz is 60 g per mol. So, 60 g quartz means 1 mole quartz.

Explanation:

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Fill in the blank for the following statement: ________ is the movement of rocks, sand, or materials from a place?
Verizon [17]

Answer:

Erosion.

Explanation:

It can be as small as a grain of sand or as large as a boulder. Sediment moves from one place to another through the process of erosion. Erosion is the removal and transportation of rock or soil. Erosion can move sediment through water, ice, or wind.

3 0
3 years ago
Read 2 more answers
PLEASE HELPPP. I'LL AWARD BRAINLIEST !!!!!
dsp73

Answer:

the properties of catalyst are

it remains unchanged after chemical reactions

it accelerate or deaccelerate the reaction without taking part in it

Explanation:

they are used to convert raw materials into useful one

catalyst are integral in making plastics

5 0
3 years ago
State whether the following statements are true or false. If false, explain why. (a) A reaction stops when equilibrium is reache
Elden [556K]

Answer:

(a) False;

(b) False;

(c) False;

(d) True.

Explanation:

(a) When equilibrium is reached, the forward reaction rate becomes equal to the reverse reaction rate, that's why the molarity of each species remains constant, but reactions don't stop.

(b) According to the principle of Le Chatelier, an increase in molarity of either reactants or products would lead to a disturbance of equilibrium. This disturbance would lead to the shift of equilibrium towards the side which would minimize such a disturbance.

(c) Equilibrium constant is only temperature-dependent, it's independent of molarity, pressure, volume etc. of any species present in the reaction.

(d) The greater the initial molarity of reactants, the more products can be formed, e. g., since the ratio of products to reactants should be kept constant, the larger the amount of reactants, the greater the amount of products formed to keep a constant ratio.

7 0
3 years ago
View the diagram below
Sauron [17]

Answer : The correct answer is the Bonds were broken on the reactants and new bonds were formed on the products.

Explanation :

In the chemical reaction, some substances react together are called reactant and the substance are formed are called product.

During the chemical reaction, the atoms of reactants rearranged to make products. There are on atoms are added or taken away in the reaction. This is known as the conservation of atoms.

For example : carbon atom react with the oxygen to form carbon dioxide.

C+O_2\rightarrow CO_2

From the given diagram, we conclude that the arrangement of molecules are different on both side of the mixture of reaction.

On the reactant side, the red molecules bonded with red molecule and the black molecule with white molecules. On the other hand i.e product side, the red molecule bonded with black molecule and white molecule bonded with red molecules. The molecular arrangement are different on both side of the reaction mixture.

Therefore, the correct answer is the Bonds were broken on the reactants and new bonds were formed on the products.


6 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
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