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Usimov [2.4K]
3 years ago
14

Which element is chemically similar to chlorine?

Chemistry
1 answer:
ad-work [718]3 years ago
6 0

bromine

Explanation:

halogens are a group of elemnts simlar to eachother

flourine, chlorine, and bromine

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Identify a predator from a Himalayan Mountain ecosystem. How else can this animal be classified in terms of feeding relationship
postnew [5]

Answer:

A snow leopard  is one of the top consumers in the Himalayas who lives in dens that are close to somewhere to look down to watch its prey. They also use it's large paws to climb up slopes and snow and a long tale to balance on thin spaces to catch markhhors.

Explanation:

<h3>(:</h3>
6 0
2 years ago
When a solution of ammonium sulfate is added to a
Artist 52 [7]

Answer:

Explanation:

(NH4)2SO4+Pb(NO3)2...............PbSO4+2NH4NO3

7 0
3 years ago
Mixtures are classified as homogeneous and heterogeneous according to _____.
Dafna1 [17]

Answer:

i believe its the 1st option

8 0
2 years ago
Suppose you needed a 0.0250 M sodium thiosulfate solution to conduct 6 titrations. Explain how you would make up this solution u
AysviL [449]

Explanation:

According to the law of dilution,

       M_{1}V_{1} = M_{2}V_{2}

The given data is as follows.

   M_{1} = 0.4782,     V_{1} = ?

    M_{2} = 0.025 ,    V_{2} = 250 mL

Hence, we will calculate the value of V_{1} as follows.

              V_{1} = \frac{M_{2} \times V_{2}}{M_{1}}

                          = \frac{0.025 \times 250}{0.4782}

                          = 13.07

Thus, we can conclude that we need 13.07 mL 0.4782 M sodium thiosulfate solution using pipette.

4 0
3 years ago
Given that Delta.G for the reaction below is –957.9 kJ, what is Delta.Gf of H2O?
Alexxandr [17]

Answer:

6ΔG°(f) H₂O = -229 Kj/mol

Explanation:

                    4NH₃(g)          +      5O₂(g)       =>        4NO(g)           +     6H₂O(g)

ΔG°(f) 4mol(-16.66Kj/mol) | 5mol(0Kj/mol) || 4mol(+86.71Kj/mol) | 6ΔG°(f) H₂O

Hess's Law

ΔG°(Rxn) = ∑ΔG°(f) Products - ∑ΔG°(f) Reactants

-957.9 Kj = [(4mol(+86.71Kj/mol)) + 6ΔG°(f) H₂O(g)] - [4mol(-16.66Kj/mol) + 5mol(0Kj/mol)]

-957.9 Kj = [4(86.7)Kj + 6ΔG°(f) H₂O] - [4(-16.66)Kj] = 346.84Kj + 6ΔG°(f) H₂O + 66.64Kj

ΔG°(f) H₂O = ((-957.9 - 346.84 -66.64)/6)Kj =  -228.56 Kj ≅ -228.6 Kj*

*Verified with Standard Heat of Formation Table

8 0
3 years ago
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