You have a particle of length 68 nm. What is this in meters? ( need help asap)

8.03 Lab Report Data Answers (I only need the Data as I do not have the time to get the materials and do it myself but I don't n

Gekata [30.6K]

Answer:

subject?

Explanation:

7 0

3 years ago

Na3AsO4 is a salt of a weak base that can accept more than one proton. If 18.4 g of Na3AsO4 is dissolved in water to make 250mL

harina [27]

Answer:

0.266 moles of Na⁺

Explanation:

First step we dissociate the salt:

Na₃AsO₄ → 3Na⁺ + AsO₄⁻³

From 1 mol of sodium arsenate, we must have 3 moles of sodium cation and 1 mol of arsenate.

We determine the moles of salt:

18.4 g . 1 mol/ 207.89 g = 0.0885 moles of salt.

We apply the followring rule of three:

1 mol of salt has 3 moles of Na⁺

0.0885 moles of salt may have (0.0885 . 3) / 1 = 0.266 moles of Na⁺

4 0

3 years ago

The reaction of NH3 and O2 forms NO and water. The NO can be used to convert P4 to P4O6, forming N2 in the process. The P4O6 can

posledela

Answer : The mass of $PH_3$ produced from the reaction is, 0.651 grams.

Explanation :

The chemical reactions used are:

(1) $4NH_3+5O_2\rightarrow 4NO+6H_2O$

(2) $6NO+P_4\rightarrow P_4O_6+3N_2$

(3) $P_4O_6+6H_2O\rightarrow 4H_3PO_4$

(4) $4H_3PO_4\rightarrow PH_3+3H_3PO_4$

First we have to calculate the moles of $NH_3$

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of $NH_3$ = 17 g/mol

$\text{Moles of }NH_3=\frac{1.95g}{17g/mol}=0.115mol$

Now we have to calculate the moles of $NO$

From the balanced chemical reaction 1, we conclude that:

As, 4 moles of $NH_3$ react to give 4 moles of $NO$

So, 0.115 moles of $NH_3$ react to give 0.115 moles of $NO$

Now we have to calculate the moles of $P_4O_6$

From the balanced chemical reaction 2, we conclude that:

As, 6 moles of $NO$ react to give 1 moles of $P_4O_6$

So, 0.115 moles of $NO$ react to give $\frac{0.115}{6}=0.0192$ moles of $P_4O_6$

Now we have to calculate the moles of $H_3PO_4$

From the balanced chemical reaction 3, we conclude that:

As, 1 moles of $P_4O_6$ react to give 4 moles of $H_3PO_4$

So, 0.0192 moles of $P_4O_6$ react to give $0.0192\times 4=0.0768$ moles of $H_3PO_4$

Now we have to calculate the moles of $PH_3$

From the balanced chemical reaction 4, we conclude that:

As, 4 moles of $H_3PO_4$ react to give 1 moles of $PH_3$

So, 0.0768 moles of $H_3PO_4$ react to give $\frac{0.0768}{4}=0.0192$ moles of $PH_3$

Now we have to calculate the mass of $PH_3$

$\text{ Mass of }PH_3=\text{ Moles of }PH_3\times \text{ Molar mass of }PH_3$

Molar mass of $PH_3$ = 33.9 g/mole

$\text{ Mass of }PH_3=(0.0192moles)\times (33.9g/mole)=0.651g$

Therefore, the mass of $PH_3$ produced from the reaction is, 0.651 grams.

6 0

3 years ago

Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and

blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g; Hydrogen H = 2.24 g Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

$=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon$

For Hydrogen:

$=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen$

For Oxygen:

$=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen$

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon$

For Hydrogen:

$\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen$

For Oxygen:

$\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen$

Thus, the empirical formula is HCO₂

4 0

3 years ago

The specific heat of water is 4.18 J/(g * °C) How much heat, in kilojoules, must be added to 250 g of water to increase the temp

katovenus [111]

You need to use the equation q=mcΔT.
q=the heat absorbed or released
m= the mass of the sample (in this case 250g)
c=the specific heat of the sample (in this case 4.18J/g°C)
ΔT=the change in temperature (in this case 5°C)
When you plug every thing in you should get q=5225J or 5.225kJ

I hope this helps. Let me know if anything is unclear.

4 0

3 years ago