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3 years ago

A tax that increases in proportion to increase in income is known as

Chemistry

1 answer:

Natalija [7]3 years ago

7 0

Im pretty sure its Taxes?

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Please help me don’t scam

rosijanka [135]

Answer:b

Explanation:

4 0

3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

5 0

3 years ago

What approximate percentage of the Earth’s freshwater is groundwater?

grigory [225]

the answer is c.) 30%

4 0

3 years ago

Read 2 more answers

Please help, with step by step work

natali 33 [55]

$\qquad$ ☀️ $\pink{\bf{ {Answer = \: \: 85.57g }}}$

Molar mass of $\bf Cu_2O$

$\qquad$ $\twoheadrightarrow\sf 63.546 \times 2 +16$

$\qquad$ $\pink{\twoheadrightarrow\bf 143.092 g}$

As we know–

1 mol = $\bf 6.02×10^{23}$ formula units

1 mol $\bf Cu_2O$ = 143.092 g = $\bf 6.02×10^{23}$ formula units

Henceforth –

$\bf 3.60×10^{23}$ formula units $\bf Cu_2O$ –

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×10^{23 }}{6.02×10^{23}}$

$\qquad$ $\sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23 }}}{6.02×\cancel{10^{23}}}$

$\qquad$ $\pink{:\implies\bf 85.57 g}$

5 0

2 years ago

at what temperature in celsius would a gas have volume of 13.5 L at a pressure of 0.723 atm, if it had a volume of 17.8 L at a p

Alex

Answer:

initial temperature= $-3.31^{\circ}C$

Explanation:

Assuming that the given follows the ideal gas nature;

$P_1=0.723atm$

$V_1=13.5L$

$T_1=?$

$P_2=0.612atm$

$V_2=17.8L$

$T_2=28 ^{\circ}C =273+28K=301K$

mole of gass will remain same at any emperature:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

putting all the value we get:

$T_2 =269.7K =-3.31 ^{\circ} C$

initial temperature= $-3.31^{\circ}C$

6 0

3 years ago