A substance that is dissolved in another substance is which of the following?

A 2.5 L sample of gases at STP (standard temperature and pressure is 273 k and 1.00 atm). When the temperature is raised to 273d

podryga [215]

Answer:

x = 5.9

Explanation:

6 0

3 years ago

If 1 mile = 5,280ft. how many miles are in 11,245ft

slavikrds [6]

2.13 miles

You can also check it on google

3 0

4 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

1. Convert Molecules to Moles

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

2. Convert Moles to Grams

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

4 0

3 years ago

A weather balloon is inflated to a volume of 27.9L at a pressure of 732mmHg and a temperature of 30.1?C. The balloon rises in th

masya89 [10]

Answer:

45.4 L

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 27.9 L

V₂ = ?

P₁ = 732 mmHg

P₂ = 385 mmHg

T₁ = 30.1 ºC

T₂ = -13.6 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (30.1 + 273.15) K = 303.25 K

T₂ = (-13.6 + 273.15) K = 259.55 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{732}\times {27.9}}{303.25}=\frac{{385}\times {V_2}}{259.55}$

$\frac{385V_2}{259.55}=\frac{20422.8}{303.25}$

$148225V_2=6729708.26018$

Solving for V₂ , we get:

V₂ = 45.4 L

7 0

4 years ago

For the chemical reaction, 2HCl + Ca(OH)2 ➡️ CaCl2 + 2H2O what mass of calcium hydroxide in grams is needed to produce 3.63 mol

wlad13 [49]

Answer:

2HCl (aq) + Ca

(

O

H

)

2

(aq) ---------> Ca

C

l

2

(ppt) + 2

H

2

O (aq)

let us calculate the number of moles , as per the chemical reactions;

2 moles of HCl solution reacts with one mole Calcium Hydroxide

Ca

(

O

H

)

2

One mole of HCl has mass : 36.5 g/mol, two moles of HCl will have mass, 73 g.

One mole of Ca

(

O

H

)

2

has mass 74.1 g

as per equation; 73 g of HCl reacts with 74.1 g of Ca

(

O

H

)

2

1g of HCl reacts with 74.1g / 73 of Ca

(

O

H

)

2

1g of HCl reacts with 1.015 g of Ca

(

O

H

)

2

NOW AS PER THE QUESTION MOLARITY AND VOLUME OF HYDROCHLORIC ACID IS GIVEN, IT CAN BE USED TO CALCULATE THE MASS OF HYDROCHLORIC ACID IN THE SOLUTION.

NUMBER OF MOLES of HCl ; Molarity of solution x Volume of Solution

# of moles of HCl = (0.40 mol/L ) x 350 mL

= (0.40 mol/L ) x 0.350 L = 0.14 mol

mass of HCl that makes 0.14 mol of HCl = # of moles x molar mass of HCl

mass of HCl = 0.14 mol x 36.5 g/ mol

mass of HCl = 5.11g

As per Stoichiometry , 1g of HCl reacts with 1.015 g of Ca

(

O

H

)

2

, 5.11g of HCl can react with 5.11 x 1.015 = 5.1865 g or 5.2 g of

Ca

(

O

H

)

2

.

5 0

3 years ago