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Svetlanka [38]
3 years ago
6

Can an ionic bond exist in a molecular formula?

Chemistry
1 answer:
bogdanovich [222]3 years ago
6 0

Answer:

yes

Explanation:

bark bark bark bark bark bark bork bork bork

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Physical science!!! please help
IgorLugansk [536]

Answer:

The correct answer is : No, because there are 4 hydrogen atoms on the reactants side and 2 on the products side.

Explanation:

NH_4NO_3\rightarrow N_2O+H_2O

The given reaction equation is not balanced because:

  • Number of hydrogen atoms  on both sides are not equal that is 4 on reactants side and 2 on products side.
  • Number of oxygen atoms on both sides are not equal that is 3 on reactants side and 2 on products side.

In a balanced chemical equation number of atoms of each elements are equal on both sides.

So, the balanced chemical equation will be:

NH_4NO_3\rightarrow N_2O+2H_2O

7 0
4 years ago
Use the periodic table in the tools bar to answer these questions. How many moles of AgNO3 are present in 1.50 L of a 0.050 M so
IgorC [24]
M= moles de soluto / litros de solucion 
moles de soluto = M. litros de solucion 

Moles de soluto = 0.050 M x 1.50 L = 0.075 moles de AgNo3
4 0
3 years ago
Read 2 more answers
Use Avogadro's number, 6.02E23, to calculate the number
Zepler [3.9K]

Answer:

<h2>2.408 × 10²¹ is the correct answer!!</h2>
8 0
3 years ago
A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature rose b
dezoksy [38]

Answer:

872.28 kJ/mol

Explanation:

The heat released is:

ΔH = C*ΔT

where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:

ΔH =  9.47*5.72 = 54.1684kJ

A quantity of 1.922 g of methanol in moles are:

moles = mass / molar mass

moles = 1.992/32.04 = 0.0621 mol

Then the molar heat of combustion of methanol is:

ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol

5 0
4 years ago
How much of 45 grams of pu 234 remains after 27 hours if it’s half life is 4.98 hours
nadya68 [22]

Answer:

1.07 g

Explanation:

Half-life of Pu-234 = 4.98 hours

Initially present = 45 g

mass remains after 27 hours = ?

Solution:

Formula

         mass remains = 1/ 2ⁿ (original mass) ……… (1)

Where “n” is the number of half lives

To find "n" for 27 hours

                          n = time passed / half-life . . . . . . . .(2)

put values in equation 2

                          n = 27 hr / 4.98 hr

                          n = 5.4

Mass after 27 hr

Put values in equation 1

          mass remains = 1/ 2ⁿ (original mass)

          mass remains = 1/ 2^5.4 (45 g)

          mass remains = 1/ 42.2 (45 g)

          mass remains =  0.0237 x 45 g

          mass remains =  1.07 g

6 0
4 years ago
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