If the statement is true, write true. If the statement is false, change the underlined word or words to make the statement true.

I don't know how to do this can I get the answers plz it's due in 1 hour

katrin [286]

Okay, so even if I just gave you the answers, your teacher needs work on it too so it'll be easier/better if I just explain how to do it.
Basically, both sides need to have the same number of molecules. To do this, we make charts. This is the first side of number one:
Na - 1
Mg- 1
F - 2
The subscript gives F two molecules, and the other ones only each have one. This is the second side:
Na- 1
Mg- 1
F- 1
So they're not equal. To fix this, we add coefficients. These are numbers that are going to appear in the front of each compound/element and changes the number of molecules of the WHOLE compound/element. We need two F on the second side, so we'll put a coefficient of 2 in front of NaF. The new chart for the second side is this:
Na- 2
Mg- 1
F- 2
Now we've fixed the F, but now Na is off! So let's go to the first side again and see what we can do. We can put a 2 in front of the Na. The new chart is this:
Na- 2
Mg -1
F- 2
Now both sides are the same. The full new equation is:
2Na + MgF(sub2) = 2NaF + Mg
Basically, do this for all of them. Feel free to ask more questions.

4 0

2 years ago

Read 2 more answers

A 1.52 g sample of KCIO3 is reacted according to the balanced equation below. How many liters of O2 is produced at a pressure of

ipn [44]

Answer:

0.486 L

Explanation:

Step 1: Write the balanced reaction

2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)

Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃

The molar mass of KCIO₃ is 122.55 g/mol.

1.52 g × 1 mol/122.55 g = 0.0124 mol

Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃

The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol

Step 4: Calculate the volume corresponding to 0.0186 moles of O₂

0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L

7 0

2 years ago

What happens to the particles in a substance as it freezes?

Ierofanga [76]

they move around and eventually come and join to make a solid

8 0

3 years ago

Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromi

GuDViN [60]

<u>Answer:</u> The mass of bromine reacted is 160.6 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol$

The chemical equation for the reaction of aluminium and bromide follows:

$2Al+3Br_2\rightarrow 2AlBr_3$

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = $\frac{3}{2}\times 0.670=1.005mol$ of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

$1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g$

Hence, the mass of bromine reacted is 160.6 grams.

5 0

2 years ago

What happens when piece of sodium is exposed to moist air?

I am Lyosha [343]

Answer:

Sodium reacts with the oxygen in air to form sodium oxide, and traces of yellowish sodium peroxide. ... Probably, the reaction with atmospheric oxygen will be faster, because I have never seen sodium metal turning whitish on exposure, though it does lose its lustre and and reactivity (just the surface).

5 0

2 years ago