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3 years ago

Um can someone just help me with the first question so I have an idea of what to do...please

Chemistry

1 answer:

Mice21 [21]3 years ago

4 0

look at the clues by it and try not to trust the links they trying to give u...

but i kinda dont know myself any periodic table i can look at?

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_H2 SiCl2 + _H2O −−→ _H8 Si4O4 + _HCl

Serjik [45]

Answer:

Explanation:

pls mark brainliest right

6 0

3 years ago

How do i balance the first chemical reaction above?? plz plz help

soldier1979 [14.2K]

Answer:

2K +F₂→ 2KF

Explanation:

When we balance an equation, we are trying to ensure that the number of atoms of each element is the same on both sides of the arrow.

On the left side of the arrow, there is 1 K atom and 2 F atoms. On the right, there is 1 K and 1 F atom.

Since the number of K atoms is currently balanced, balance the number of F atoms.

K +F₂→ 2KF

Now, that the number of F atoms is balanced on both sides, check if the number of K atoms are balanced.

Left

K atoms: 1

F atoms: 2

Right

K atoms: 2

F atoms: 2

The number of K atoms is not balanced.

2K +F₂→ 2KF

Left

K atoms: 2

F atoms: 2

Right

K atoms: 2

F atoms: 2

The equation is now balanced.

4 0

2 years ago

Lead has a density of 11.29 g/mL. If you had 186 g of lead, what would its volume be?

Varvara68 [4.7K]

I think you just do 11.29 multiplied by 186 since to find density you divide mass by volume. So 186 divided by x is 11.29. So in conclusion the volume would be 2,099.94mL

8 0

3 years ago

Which two factors that influence weather are caused by the uneven heating of

just olya [345]

B, sorry if i am wrong.

8 0

2 years ago

Read 2 more answers

How many grams of helium must be released to reduce the pressure to 65 atmatm assuming ideal gas behavior? Express the answer in

lyudmila [28]

The question is incomplete, here is the complete question:

How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.

Answer: The mass of helium released is 1.6 grams

Explanation:

We are given:

Mass of helium in the cylinder = 5.209 g

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{w}{M}RT$

where,

P = Pressure of the gas = 65 atm

V = Volume of the gas = 334 mL = 0.334 L (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = ?

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $23^oC=[23+273]K=296K$

Putting values in above equation, we get:

$65atm\times 0.334L=\frac{w}{4g/mol}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 296K\\\\w=\frac{65\times 0.334\times 4}{0.0821\times 296}=3.573g$

Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g

Hence, the mass of helium released is 1.6 grams

5 0

3 years ago