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1 year ago

How many significant figures does 7,874.9640 s have? Submit Question

Chemistry

2 answers:

julia-pushkina [17]1 year ago

8 0

Answer:

8 significant figures

Explanation:

All numbers (that are not zero) are always significant. Because there is a zero at the end of the number after the decimal place, this makes this zero also significant. This zero is significant because it makes the number more precise. Therefore, all of the numbers in this number are significant.

Bond [772]1 year ago

4 0

Answer:

There are 8 sig figs

Explanation:

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C= ( F - 32) / 1.8<br> 86 F = ?

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2 years ago

If the mass of an object is 60.0 grams and the density of the object is 15.0 g/mL, what is the volume of the object? The formula

laila [671]

Answer:

168.56 mL

Explanation:

density = mass/volume, use basic algebra skills to replace the volume and solve

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2 years ago

Consider the reaction. 2 HBr(g) ¡ H2(g) + Br2(g) a. Express the rate of the reaction in terms of the change in concentration of

Studentka2010 [4]

Answer :

(A) The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

(B) The average rate of the reaction during this time interval is, 0.00176 M/s

(C) The amount of Br₂ (in moles) formed is, 0.0396 mol

Explanation :

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The given rate of reaction is,

$2HBr(g)\rightarrow H_2(g)+Br_2(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }HBr=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Rate of disappearance of }H_2=+\frac{d[H_2]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{d[Br_2]}{dt}$

The rate expression will be:

$Rate=-\frac{1}{2}\frac{d[HBr]}{dt}=+\frac{d[H_2]}{dt}=+\frac{d[Br_2]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{d[HBr]}{dt}$

$\text{Average rate}=-\frac{1}{2}\frac{(0.512-0.600)M}{(25.0-0.0)s}$

$\text{Average rate}=0.00176M/s$

The average rate of the reaction during this time interval is, 0.00176 M/s

As we are given that the volume of the reaction vessel is 1.50 L.

$\frac{d[Br_2]}{dt}=0.00176M/s$

$\frac{d[Br_2]}{15.0s}=0.00176M/s$

$[Br_2]=0.00176M/s\times 15.0s$

$[Br_2]=0.0264M$

Now we have to determine the amount of Br₂ (in moles).

$\text{Moles of }Br_2=\text{Concentration of }Br_2\times \text{Volume of solution}$

$\text{Moles of }Br_2=0.0264M\times 1.50L$

$\text{Moles of }Br_2=0.0396mol$

The amount of Br₂ (in moles) formed is, 0.0396 mol

8 0

2 years ago

What happens when the temperature of a solution increases?

bazaltina [42]

Answer:

<h3>when the temperature of a solution increases then,solubility of solid solute increases...</h3>

4 0

3 years ago

Read 2 more answers

Chlorine gas is added to a large flask to a pressure of 1.85 atm, at a temperature of 322 K. Phosphorus is added, and a reaction

AVprozaik [17]

Answer:

The volume of the flask is<u> 20.245 litres .</u>

Explanation:

We are given with following information-

$PV=nRT$ -------- 1

where R = $0.0821L.atom/mole.K$

Molar mass of $PCl_5=208.22g/mole$

The given chemical equation is -

$2P _(_s_)+5Cl_2 _(_g_)\rightarrow2PCl_5 _(_s_)$ --------- 2

Now , calculation -

Mass of $PCl_5$ formed = 118g

Molar mass of $PCl_5$ = $208.22g/mole$

Mole = $\frac{mass (g)}{molar mass}$

Therefore , moles of $PCl_5$ formed = $\frac{118}{208.22}$

From equation 2 , we get to know that ,

2mole $PCl_5$ formed from 5 mole $Cl_2 _(_g_)$

Therefore , $\frac{118}{208.22}$ mole $PCl_5$ formed from $\frac{5}{2} \times\frac{118}{208.22}$ mole $Cl_2 _(_g_)$

Moles of $Cl_2 _(_g_)$ used = $\frac{5\times118}{2\times208.22} mole$

R= $0.0821L.atom/mole.K$

Pressure (P)= 1.85atm

Temperature (T)= 322K

Moles of $Cl_2 _(_g_)$ (n)= $\frac{5}{2} \times\frac{118}{208.22}$ moles

Applying the formula above in 1 equation , that is

PV = nRT

putting the given values -

$1.85 \times V=\frac{5}{2} \times\frac{118}{208.22}\times0.0821\times322$

V = 20.245 litres.

Hence , the volume of the flask is <u>20.245 litres . </u>

8 0

2 years ago