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Sergeu [11.5K]
3 years ago
5

Help pleaseeeeeeeeeeeeee!!?

Chemistry
2 answers:
melisa1 [442]3 years ago
4 0

Answer:

There are 146 neutrons in Uranium-238

There are 90 electrons in Thorium-234

vodka [1.7K]3 years ago
4 0
146 N in uranium and 90 E in Thorium
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After fertilization, female cones become very:<br> sticky<br> light<br> O hard
S_A_V [24]

Answer:

light

Explanation:

sorry don't now this one

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3 years ago
How far below sea level is the peak of the ridge?
mario62 [17]
I think is 1 and a half km
7 0
3 years ago
How to find the net ionic of an equation
stepan [7]
Net ionic equation can be written from complete ionic equation by canceling the spectators ion from complete ionic equation.
For example:
Balanced Chemical equation:
HClO₂(aq) + NaOH(aq) → H₂O(l) + NaClO₂ (aq)
Ionic equation:
H⁺(aq) + ClO₂⁻(aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + ClO₂⁻ (aq)
Net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The ClO₂⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
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6 0
2 years ago
The name of CuO is what
kotykmax [81]

CuO: Copper oxide


"Copper(II) oxide or cupric oxide is the inorganic compound with the formula CuO. A black solid, it is one of the two stable oxides of copper, the other being Cu₂O or cuprous oxide. As a mineral, it is known as tenorite."

5 0
3 years ago
How many grams of silver cholride will be precipitated by adding sufficient silver nitrate to react with 1500.0mL of.400M barium
Taya2010 [7]

Answer: 172.2 g of AgCl is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}={\text {Molarty}}\times {\text{Volume in L}}=0.400\times 1.5L=0.6moles

BaCl_2+2AgNO_3\rightarrow 2AgCl+Ba(NO_3)_2

According to stoichiometry:

1 mole of BaCl_2 produce = 2 moles of AgCl

Thus 0.6 moles BaCl_2 will produce =\frac{2}{1}\times 0.6=1.2 moles of AgCl

Mass of AgCl=moles\times {\text {molar mass}}=1.2mol\times 143.5g/mol=172.2g

Thus 172.2 g of AgCl is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.

4 0
3 years ago
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