All categories

3 years ago

Sub atomic particle mass

Chemistry

1 answer:

melamori03 [73]3 years ago

4 0

1. neutron
2. proton which is positive
3. electron which is negative

You might be interested in

20 Points and Brainliest if you answer in 5 mins

wariber [46]

The correct answer is D) X: FM, Y: AM

4 0

2 years ago

Jack is recording the growth of a single tree over the course of a year. What type of graph should he use to display the data?

Dmitry [639]

The answer should be the last one Line graph. I hope I helped!

5 0

3 years ago

Read 2 more answers

200 g and 5% baking soda solution must be prepared. How much soda (in grams) and how much water (in milliliters) will be needed?

Vanyuwa [196]

Answer:

190 mL of water

Explanation:

5% solution means that 5 g of substance in 100 g of solution.

So, if we have 5 g of substance, and it has to be 100 g solution, than we need 100-5 = 95 g water.

100 g solution --- 5 g soda--- 95 g water

200g solution --- 10 g soda ---190 g water

For 200 g of 5% solution, we will need 10 g of soda and 200-10= 190 g of water.

Density of water = 1 g/mL

190 g * 1ml/g = 190 ml water

5 0

3 years ago

Vinegar contains carbon, hydrogen, and oxygen with percent masses of 40.01% C and 6.70% H. If the molar mass of vinegar is about

oee [108]

Answer: The molecular formula will be $C_2H_4O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 40.01 g

Mass of H = 6.70 g

Mass of O = 100-(40.01+ 6.70) = 53.29 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.01g}{12g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.70g}{1g/mole}=6.70moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.29g}{16g/mole}=3.33moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{6.70}{3.33}=2$

For O = $\frac{3.33}{3.33}=1$

The ratio of C : H: O= 1: 2: 1

Hence the empirical formula is $CH_2O$

The empirical weight of $CH_2O$ = 1(12)+2(1)+1(16)= 30 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2$

The molecular formula will be= $2\times CH_2O=C_2H_4O_2$

6 0

3 years ago

Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.65 g of N2H4 reacts with

White raven [17]

Answer: The percent yield of the reaction is 17.41 %.

Explanation:

For $N_2H_4$

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $N_2H_4$ = 2.65 g

Molar mass of $N_2H_4$ = 32.04 g/mol

Putting values in above equation, we get:

$\text{Moles of }N_2H_4=\frac{2.65g}{32.04g/mol}=0.0827mol$

For $N_2$

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

$PV=nRT$

where,

P = pressure of the gas = 1 atm

V = Volume of gas = 0.350 L

n = Number of moles = ?

R = Gas constant = $0.0820\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the gas = 273 K

Putting values in above equation, we get:

$1.00atm\times 0.350L=n\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\n=0.0144mol$

Now, to calculate the experimental yield of $N_2$ , we use the equation:

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

We are given:

Moles of nitrogen gas = 0.0144 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$0.0144mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=0.4032g$

Experimental yield of nitrogen gas = 0.4032 g

For the given chemical equation:

$N_2H_4(aq.)+O_2(g)\rightarrow N_2(g)+2H_2O(l)$

By Stoichiometry of the reaction:

1 mole of $N_2H_4$ produces 1 mole of nitrogen gas.

So, 0.0827 moles of $N_2H_4$ will produce = $\frac{1}{1}\times 0.0827=0.0827mol$ of nitrogen gas.

Now, to calculate the theoretical yield of nitrogen gas, we use equation 1:

Moles of nitrogen gas = 0.0827 mol

Molar mass nitrogen gas = 28 g/mol

Putting values in above equation, we get:

$0.0827mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=2.3156g$

Theoretical yield of nitrogen gas = 2.3156 g

To calculate the percentage yield of nitrogen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of nitrogen gas = 0.4032 g

Theoretical yield of nitrogen gas = 2.3156 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{0.4032g}{2.3156g}\times 100\\\\\% \text{yield of nitrogen gas}=17.41\%$

Hence, the percent yield of the reaction is 17.41 %.

8 0

2 years ago