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2 years ago

1.0 gallon to mmN.B 1 gallon=3.8L

Chemistry

1 answer:

I am Lyosha [343]2 years ago

4 0

Answer:

Gallon (US) to Milliliter Conversion Table

Gallon (US) [gal (US)] Milliliter [mL]

1 gal (US) 3785.411784 mL

2 gal (US) 7570.823568 mL

3 gal (US) 11356.235352 mL

5 gal (US) 18927.05892 mL

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How many oxygen atoms are there in 0.25 mole of CO2?

Anna35 [415]

Answer:

It is known that 1 mol of a molecule contains 6.023×1023 6.023 × 10 23 number of molecules. So, 0.25 moles of CO2 C O 2.

3 0

2 years ago

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

___________________________

$\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}$

4 0

2 years ago

100.mL of a .795 M solution of KBr is diluted to 500.mL. what is the new concentration of the solution?

kap26 [50]

Answer:

0.159 M

Explanation:

convert from mL to L then use the equation:

M1V1 = M2V2

rearrange to find M2

$\frac{M1V1}{V2} = M2$

$\frac{(0.795 M)(0.100 L)}{0.500 L} = 0.159 M$

4 0

2 years ago

The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc

jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

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

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

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

Given: $-\frac{d[HBr]}{dt}]=0.301$

Putting in the values we get:

$\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}$

$+\frac{1d[Br_2]}{dt}=0.151$

Thus the rate of appearance of $Br_2$ is 0.151

8 0

3 years ago

Please help it’s due today I will give brainliest

Thepotemich [5.8K]

1st and 4th options are suitable answers, as these 2 changes are not exactly physical changes as it cant return back to original form and as well as its not cooling, so I feel its 1st and 4th options

7 0

2 years ago

1.0 gallon to mmN.B 1 gallon=3.8L​

1.0 gallon to mmN.B 1 gallon=3.8L