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Svetradugi [14.3K]

2 years ago

5

A 150 mL sample of hydrochloric acid (HCl) completely reacted with 60.0 mL of a 0.100 M NaOH solution. The equation for the reac

tion is given below. HCl + NaOH Right arrow. NaCl + H2O What was the original concentration of the HCl solution? 0.040 M 0.25 M 1.50 M 2.50 M

2 answers:

iragen [17]2 years ago

8 0

Answer:

0.040M

Explanation:

just took the review

goblinko [34]2 years ago

5 0

Answer:

0.040M

Explanation:

The following data were obtained from the question:

Volume of acid (Va) = 150mL

Volume of base (Vb) = 60mL

Molarity of base (Mb) = 0.1M

Molarity of acid (Ma) =..?

Next, the balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Finally, we shall determine the concentration of the acid, HCl as follow:

MaVa / MbVb = nA/nB

Ma x 150 / 0.1 x 60 = 1/1

Cross multiply to express in linear form

Ma x 150 = 0.1 x 60

Divide both side by 150

Ma = 0.1 x 60 / 150

Ma = 0.04M

Therefore, the concentration of the acid is 0.04M

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How many moles of oxygen are produced when 29.2 g of water is decomposed according to H20 H2 + O2

Katarina [22]

8.1moles

Explanation:

Given parameters:

Mass of water to be decomposed = 29.2g

Unknown:

Number of moles of oxygen.

Solution:

To solve this problem, we first write the balanced reaction equation :

2H₂O → 2H₂ + O₂

Now convert the given mass of the water to number of moles;

Number of moles of water = $\frac{mass}{molar mass}$

Molar mass of water = 2(1) + 16 = 18g/mol

Number of moles of water = $\frac{29.2}{18}$ = 16.2moles

From the balanced reaction equation:

2 moles of water produced 1 mole of oxygen gas;

16.2 mole of water will produce $\frac{16.2}{2}$ = 8.1moles of oxygen gas

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

5 0

2 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)?<br><br> Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

6 0

2 years ago

Cart A has a mass of 24 kg and is being pulled with a force of 32 N. Cart B has a mass of

horsena [70]

Cart b because it has greater force and higher mass

4 0

2 years ago

If the reactants are 10 kg what is the mass of the products

nlexa [21]

If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.

It is due to the law of conservation of mass.

Mass can neither be created nor be destroyed.

7 0

2 years ago

Calculate the volume of the gas, in liters, if 1.75 mol has a pressure of 1.28 atm at a temperature of -7 ∘C

Alexandra [31]

Answer:

A sample of an ideal gas has a volume of 2.21 L at 279 K and 1.01 atm. Calculate the pressure when the volume is 1.23 L and the temperature is 299 K.

You need to apply the ideal gas law PV=nRT

You have the pressure, P=1.01 atm

you have the volume, V = 2.21 L

The ideal gas constant R= 0.08205 L. atm/ mole.K at 273 K

find n = PV/RT = (1.01 atm x 2.21 L / 0.08205 L.atm/ mole.K x 273 K)

n= 0.1 mole, Now find the pressure for n=0.1 mole, T= 299K and

L=1.23 L

P=nRT/V= 0.1mole x 0.08205 (L.atm/ mole.K x 299 k)/ 1.23 L

= 1.994 atm

Explanation:

6 0

2 years ago