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Levart [38]
3 years ago
8

1,670,000,000 in scientific notation pleaseeee

Chemistry
2 answers:
noname [10]3 years ago
6 0

Answer:

1.67 × 10^9

Explanation:

Alchen [17]3 years ago
4 0
1.67*10^9 is the answer, because with an exponent the number on the left must be less than 10. Multiplying that number by 10^9 moves the decimal to the right 9 times, adding 7 zeroes.
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Covalent solutes are considered non-electrolytes. What does this mean for the conductivity of the solution? A) Non-electrolytes
iogann1982 [59]

Electrolytes are those which dissociates in solution and produces ions.

Ions can carry current,so Electrolytes conduct electiricity.

And non electrolytes are those which do not dissociate in solution and doesnt produce ions.

Since non electrolytes do not produce ions they cannot conduct electricity.

<u>Hence the right option is:</u>

B) Non-electrolytes dissolve and do not dissociate in water providing no charged ions to conduct electricity.  

5 0
3 years ago
Read 2 more answers
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4 0
4 years ago
Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature
weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}     ------>NO_{(g)}

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

\delta H^0__{R,T_2} = \delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'

where:

\delta H^0__{R} = enthalpy of reaction

{\delta C_p(T')} = the difference in the heat capacities of the products and the reactants.

∴

\delta H^0__{R,435K} = \delta H^0__{R,298.15K} + \int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'

= 1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

\delta H^0__{R,435K} ≅ 91383 J

6 0
4 years ago
A student was asked to prepare 500.0 mL of 6.0 M NaOH. The student measured 120.0 g of NaOH and placed it in a 1000 mL beaker. T
Morgarella [4.7K]

Answer:

The concentration of the solution will be much lower than 6M

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

n= CV

n = number of moles m/M( m= mass of solid, M= molar mass of compound)

C= concentration of substance

V= volume of solution

m=120g

M= 40gmol-1

V=500ml

120/40= C×500/1000

C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

7 0
3 years ago
Giúp mình 2 câu này với ạ
alisha [4.7K]

Answer:it is wrong answer

Explanation:estro man

5 0
2 years ago
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