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2 years ago

McCray, Shy'Anne

Chemistry

1 answer:

dolphi86 [110]2 years ago

3 0

Answer:

The temperature of soup will decrease.

The soup will get as cold as the room temperature.

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How many moles are there in 3.45 moles of Carbon

VladimirAG [237]

Answer:

It would be 151.832775 because one mole is 44.0095*3.45 i hope this helps!

Explanation:

7 0

2 years ago

The characteristic functional groups of a protein are _____.

V125BC [204]

Answer:

$\boxed{\text{amide groups}}$

Explanation:

A protein is a long chain of amino acids linked together by amide groups.

The general structure is

$\rm \left[-NHCHR-\underbrace{\hbox{CO-NH}}_{\hbox{amide group}}-CHRCO-\right]_{n}$

4 0

2 years ago

Calculate the density of nitrogen gas, in grams per liter, at stp.

lesantik [10]

Standard temperature is 273 K

Standard pressure is 1 atm

We use the ideal gas equation to find out density of nitrogen gas in g/L

Ideal gas equation:

$PV = nRT\\ PV = (\frac{Mass}{Molar mass)}RT\\ P(Molar mass) = (\frac{Mass}{Volume})RT\\ \frac{Mass}{Volume}=\frac{P(molar mass)}{RT} \\ Density = \frac{P(Molar mass)}{RT}$

Molar mass of $N_{2} = 28 g/mol$

Pressure = 1 atm

Temperature = 273 K

$Density = \frac{(1atm)(28 g/mol)}{(0.08206 \frac{L.atm}{mol.K})(273 K)}$

= 1.25 g/L

Therefore, density of nitrogen gas at STP is 1.25 g/L

4 0

2 years ago

_____ Na3PO4 + _____ KOH arrow _____ NaOH + ___________K3PO4 <br> How do I balance this equation?

soldi70 [24.7K]

The answer is na3po4+3koh=3naoh+k3po4

6 0

2 years ago

How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci

bezimeni [28]

Answer:

Approximately $0.08\; \rm mol$ , assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure: $25\; \rm ^{\circ}C$ and $P = 101.325 \; \rm kPa$ .

The question states that the volume of this gas is $V = 2\; \rm dm^{3}$ .

Convert the unit of all three measures to standard units:

$\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}$ .

$\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}$ .

$\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}$ .

Look up the ideal gas constant in the corresponding units: $R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}$ .

Let $n$ denote the number of moles of this gas in that $V = 2\; \rm dm^{3}$ . By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

$P \cdot V = n \cdot R \cdot T$ .

Rearrange this equation and solve for $n$ :

$\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}$ .

In other words, there is approximately $2\; \rm mol$ of this gas in that $V = 2\; \rm dm^{3}$ .

6 0

2 years ago