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IRINA_888 [86]
3 years ago
7

How many significant digits are in the number 1.0 × 10^3?​

Chemistry
2 answers:
Arturiano [62]3 years ago
7 0
Answer: 1000
1•10^3 add three 0’s
Galina-37 [17]3 years ago
6 0
121837481817381813727-
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8 points
iVinArrow [24]

Answer:

Burning Fossil Fuels

Mining

Using Fertilizers

6 0
3 years ago
A 50.51g sample of a compound made from phosphorus and chlorine is decomposed. Analysis of the products showed that 11.39 g of p
Alchen [17]

<u>Ans: P = 22.5% and Cl = 77.5%</u>

<u></u>

<u>Given:</u>

Total Mass of sample containing P and Cl = 50.51 g

Mass of P produced = 11.39 g

<u>To determine:</u>

Mass % of P and Cl

<u>Explanation:</u>

Mass % of a given element in a total mass is generally expressed as:

Mass % of element = [mass of element/total mass]*100

Here,

Total mass = mass of P + mass of Cl

mass of Cl = Total - mass of P = 50.51-11.39 = 39.12 g

% P = [11.39/50.51]*100 = 22.5%

%Cl = [39.12/50.51]*100 = 77.5%

5 0
3 years ago
Read 2 more answers
5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points
labwork [276]

Answer: (A) H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B) H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of H_{3}O^{+} in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

       Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)

 NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)

Hence, net ionic equation will be as follows.

       H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B)  When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

     NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)

The net ionic equation is as follows.

        H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

7 0
3 years ago
What is the k expression for this reaction? plz help !
just olya [345]

Answer:

K = [HI]² / [H₂] [I₂]

Explanation:

To write the expression of equilibrium constant, K, it is important that we know how to obtain the equilibrium constant.

The equilibrium constant, K for a given reaction is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient. Thus, the equilibrium constant is written as follow:

K = [Product] / [Reactant]

Now, we shall determine the equilibrium constant for the reaction given in the question above. This can be obtained as illustrated below:

H₂(g) + I₂(g) —> 2HI (g)

K = [HI]² / [H₂] [I₂]

6 0
3 years ago
Convert to grams<br> 0.100 moles of Co2
omeli [17]
One mole of C= 12 grams
two moles of O =32
so one mole of CO^2 is44 grams
.1 moles or 1/10 moles of 44 grams is 4.4 grams
3 0
3 years ago
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