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2 years ago

Can someone please help me with these three questions, it will soon be due.

Chemistry

1 answer:

Mice21 [21]2 years ago

8 0

YOU also better mark me BRAINLIEST if someone else answers!!!

volume = 100 mL

density = 1.02 g / mL

mass = density x volume

= 1.02 x 100
= 102 g

Cp = 4.03 J / oc g
dT = 28.7 - 23.5 = 5.2 oC
Q = m Cp dT
Q = 102 x 4.03 x 5.2
Q = 2138 J
Q = 2.14 kJ

heat absorbed = 2.14 kJ

We can calculate the moles of HC1 present in 50 ml 1m solution by using the formula, moles = volume of solution in L*molarity
moles = 50/1000 = .05 , L*1 M = .05 Moles of HCL.
The same can be applied for calculation of moles fo Na OH of 50 ml, 1 M solution. (50 ml NaOH = .o5 L NaOH)
Then use the formula = volume of solution / 1000*moles.wt,
=1.825 gram.
moles=mass in grams /moles.wt= .05 moles of HCI
calculate moles of water from this reaction.
HC1+NaOH----->NACl+h20 (Balanced Equation)
.5+.5----> .5+.5
.05 moles of water
Delta H=q/n.
n=.05 to calculate the Delat H.

3) I don't understand it, sorry

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Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

7 0

3 years ago

Which ph range is ideal for nitrogen in corn crops

labwork [276]

The desirable soil pH for corn ranges from 5.8 to 7.0.

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3 years ago

How many moles of SnF2 will be produced along with 48 grams of H2

wolverine [178]

Answer:

24 moles

Explanation:

given data

$H_2$ = 48 grams

to find out

how many moles of $SnF_2$ will be produced

solution

we know that Stoichiometry is here as

$Sn+2HF\rightarrow SnF_2+H_2$ ......................1

so we get here mole of solute $H_2$ that is express as

mole = $\frac{\text{given mass}}{\text{Molar Mass}}$ ..........................2

put here value for $H_2$

mole = $\frac{48}{2}$

mole of $H_2$ = 24 mole

and

we can say that here as per equation 1 that one mole of $H_2$ is go with one mole of $SnF_2$

so here $H_2$ 24 mole is go with 24 moles mole of $SnF_2$

so we can say $SnF_2$ will be produced 24 moles

5 0

3 years ago

An aqueous solution at 27°c contains 3.6 g of a protein in a 200 ml sample. the osmotic pressure is 0.0203 atm. what is the mola

umka21 [38]

In order to calculate the molar mass of the protein, we may manipulate the ideal gas equation:
PV = nRT, where n is the number of moles. We also know that:
n = m / Mr, where m is mass and Mr is molecular weight
Thus,
Mr = (mRT)/(PV)
Here, the mass is in grams, the temperature is in Kelvin, the pressure is in atm and the volume is in liters, so the molar gas constant is 0.082057.

Mr = (3.6 * 0.082057 * (27 + 273)) / (0.0203 * 0.2)
Mr = 21,828 g/mol
Thus, the Mr of the protein is 2.18 x 10⁴ g/mol

8 0

3 years ago

Why are there few fossils of protists. write a paragraph response

Liono4ka [1.6K]

Answer:

protists that produce a hard outer covering have left many fossils, other protists lack hard parts, so few fossils of these organisms have been found. ... Protists in this group are called plantlike because, like plants, they contain the pigment chlorophyll in chloroplasts and can make their own food.

Explanation:

Hope it helps, some how.

3 0

2 years ago