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dimulka [17.4K]
2 years ago
8

Hydrogen gas and bromine gas react to form hydrogen bromide gas. How much heat (kJ) is released when 155 grams of HBr is formed

in this reaction? ΔH° = -72 kJ.
Chemistry
1 answer:
VashaNatasha [74]2 years ago
8 0

Answer:

-37.63KJ

Explanation:

First, we write an equation of reaction:

H + Br ——-> HBr

Now we are told that 155g of HBr is formed. We can calculate the number of moles of HBr formed as follows:

We simply divide the mass of the HBr formed by the molar mass of HBr. The molar mass of HBr is 81g/mol

The number of moles of HBr formed is thus 81/155 = 0.52 moles

The amount of heat released is thus 0.52 * (-72) = -37.63KJ

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german
16.00 g -------------- 6.02x10²³ atoms
?? g ----------------- 6.022x10²³ atoms

16.00 x (6.022x10²³) / 6.02x10²³ =

=> 16 g


7 0
2 years ago
A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3
IRINA_888 [86]

<em>The mass of a gold bar  = 1.424 x 10⁴ gram</em>

<em></em>

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

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Keywords: density, mass, volume, a gold bar

3 0
2 years ago
Read 2 more answers
Starting with 1.5052g of BaCl2•2H2O and excessH2SO4, how many grams of BaSO4 can be formed?
muminat
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba 
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
6 0
2 years ago
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According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams
gogolik [260]

Answer:

m_{HClO_3}=12.7gHClO_3

Explanation:

Hello,

Considering the reaction:

3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3

The molar masses of chlorine and chloric acid are:

M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3

Best regards.

4 0
2 years ago
What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?
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Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

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