All categories

2 years ago

Someone please help me with the two I got wrong

Chemistry

2 answers:

LenKa [72]2 years ago

8 0

Answer:

yes

Explanation:

AnnZ [28]2 years ago

7 0

Answer

yes, that is the answer

You might be interested in

How many moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide

dalvyx [7]

Answer:

5.8 moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide.

Explanation:

$2Cu+S\rightarrow Cu_2S$

Moles of copper(I) sufide = 2.9 mol

According to reaction, 1 mole of copper(I) sulfide is obtained from 2 moles of copper, then 2.9 moles of copper(I) sulfide will be obtained from :

$\frac{2}{1}\times 2.9 mol=5.8 mol$ copper

5.8 moles of copper are needed to react with sulfur to produce 2.9 moles of copper (I) sulfide.

4 0

2 years ago

Calculate the mass percent of HOCH 2CH 2OH in a solution made by dissolving 3.2 g of HOCH 2CH 2OH in 43.5g of water.

grin007 [14]

Answer:

$\%m/m=6.85\%$

Explanation:

Hello,

In this case, we are asked to compute the by mass percent for the given 3.2 g of ethylene glycol in 43.5 g of water. In such a way, since the by mass percent is computed as follows:

$\%m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%$

Whereas the solute is the ethylene glycol and the solvent the water, therefore we obtain:

$\%m/m=\frac{3.2g}{3.2g+43.5g} *100\%\\\\\%m/m=6.85\%$

Best regards.

5 0

2 years ago

A barium hydroxide solution is prepared by dissolving 2.06 g of Ba(OH)2 in water to make 32.9 mL of solution. What is the concen

navik [9.2K]

Answer: 1.36 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

To calculate the moles, we use the equation:

moles of solute= $\frac{\text {given mass}}{\text {molar mass}}=\frac{2.06g}{171g/mol}=0.0120moles$

$Molarity=\frac{0.0120\times 1000}{32.9}=0.364M$

The balanced reaction between barium hydroxide and perchloric acid:

$2HCIO_4+Ba(OH_)2\rightarrow BaCIO_4+2H_2O$

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HClO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

$n_1=1\\M_1=?\\V_1=8.50mL\\n_2=2\\M_2=0.364M\\V_2=15.9mL$

Putting values in above equation, we get:

$1\times M_1\times 8.50=2\times 0.364\times 15.9\\\\M_1=1.36M$

Thus the concentration of the acid is 1.36 M

4 0

2 years ago

(06.04 HC)

Ksju [112]

The mass of sodium chloride at the two parts are mathematically given as

m=10,688.18g
mass of Nacl(m)=39.15g

<h3>What is the mass of sodium chloride that can react with the same volume of fluorine gas at STP?</h3>

Generally, the equation for ideal gas is mathematically given as

PV=nRT

Where the chemical equation is

F2 + 2NaCl → Cl2 + 2NaF

Therefore

1.50x15=m/M *(1.50*0.0821)

1-50 x 15=m/58.5 *(1.50*0.0821)

m=10,688.18g

Part 2

PV=m'/MRT

1*15=m'/58.5*0.0821*273

m'=39.15g

mass of Nacl(m)=m'=39.15g