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2 years ago

how many grams of water will be produced if 18.15 g of naoh reacts with 14.35 g of h3po4 to give sodium phosphate and water?

Chemistry

1 answer:

DaniilM [7]2 years ago

7 0

I don't understand the elements, which ones are capital letters and which ones are lower case.

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What is the pressure in a 5.00 L tank with 49.00 grams of oxygen gas at 350 K? ___atm

ziro4ka [17]

Answer:

The right answer is "8.81 atm".

Explanation:

Given:

V = 5.00 L

Mass = 4900 g

MW = 32 g/mol

T = 350 K

Now,

Number of moles will be:

$n = \frac{Mass}{MW}$

$=\frac{49.00}{32}$

$=1.53125 \ mole$

By using the ideal gas equation, we get

⇒ $PV=nRT$

or,

⇒ $P=\frac{nRT}{V}$

By substituting the values, we get

$=\frac{1.53125\times 0.0821\times 350}{5.00}$

$=\frac{44.1}{5.00}$

$=8.81 \ atm$

4 0

2 years ago

Metallic pan is provided with a wooden plastic handle give reason

AlladinOne [14]

The metallic pan iis most likely going to be used on a stove.

The stove is heating something, and the conductive metallic pan will, well, conduct that heat throughout the entire body of the pan. Doing this will spread the heat to the handle, burning your hands.

Both wood and plastic are insulators, and they do not conduct heat or electricity. They will insulate your hands and protect them from the heat.

5 0

2 years ago

What is the mass of one iron atom in grams?

vampirchik [111]

55.8 grams Would be the answer

7 0

2 years ago

Read 2 more answers

You have a solution that is 18.5% (v/v) methyl alcohol. If the bottle contains 1.64 L of solution, what is the volume (V) in mil

Juliette [100K]

Answer:

303.4 millimeters

Explanation:

Just multiply 1.64 by 0.185. then convert that answer to millimeters

7 0

2 years ago

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0

2 years ago