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3 years ago

If the mass of a material is 78 grams and the volume of the material is 23 cm3, what would the density of the material be?

Chemistry

2 answers:

Paraphin [41]3 years ago

8 0

Density = mass / volume

Density = (78g) / (23cm^3)

Density = 3.4g/cm^3

Orlov [11]3 years ago

6 0

By definition, density is mass per volume.

$d=\frac{m}{v}$

Plugging in the given values gives:

$d=\frac{78 g}{23 cm^3} =3.3913\frac{g}{cm^3}$

The only operation done in the problem is division, and both numbers have 2 significant figures, so this would be $3.4\frac{g}{cm^3}$ with correct significant figures.

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A 4.27 g sample of a lab solution contains 1.81g of acid. What is the concentration of the solution as a mass percentage?

vaieri [72.5K]

Answer:

42.38875878%

Explanation:

i divided 4.27 g from 1.81g using a percentage calculator but im not sure if its correct

5 0

1 year ago

how many grams of solid lithium must be added to liquid after in order to obtain 15.0 L of hydrogen gas at STP?

alexdok [17]

Answer:

9.29g

Explanation:

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3 0

2 years ago

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

7 0

3 years ago

24. 00 ml of a 0. 25 m naoh solution is titrated with 0. 10m hcl. What is the ph of the solution after 24. 00 ml of the hcl has

12345 [234]

pH of the solution after 24. 00 ml of the hcl has been added is 12.87

millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00

millimoles HCl = 24.00 mL x 0.10 M = 2.40

total volume = 48.00 mL

.................................NaOH + HCl ==>NaCl + H2O

initial.........................6.00.........0............0.........0

added.....................................2.40............................

change.................... -2.40......-2.40.........+2.40.... +2.40

equilibrium.................3.60.........0..............2.40.......2.40

The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)

pOH = -log (OH^-). Then

pOH = -log (0.075)

pOH =1.1249

As we know,

pH + pOH = pKw = 14.00

pH=14-pOH

pH=14-1.1249

pH=12.87

pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.

The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.

Learn more about pH:

brainly.com/question/491373

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7 0

1 year 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