A slide 3.3 m long makes an angle of 32° with the ground. How high is the top of the slide above the ground?

The compound consists of 40% carbon and 6.71% hydrogen. Its mass of 1 liter of steam is 2.4 g. What is the formula of the compou

AveGali [126]

I think it is CH2O.

If i'm wrong srry

7 0

3 years ago

Tell me the main parts of the Earth Layers.

sukhopar [10]

Answer:

Crust (the surface)

Upper Mantle (dense magma)

Lower Mantle (less dense magma)

Outer Core (pure liquid material)

Inner Core (dense material)

Hope that helps

8 0

3 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.

azamat

Answer:

(a) $m_{gold}=7.322g$

(b)

$V_{gold}=0.379cm^3$

$V_{copper}=0.122cm^3$

(c) $\rho _{coin}=15.94g/cm^3$

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

$m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g$

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

$V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3$

$m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3} \\\\V_{copper}=0.122cm^3$

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

$\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\ \\\rho _{coin}=15.94g/cm^3$

Best regards.

3 0

3 years ago

Read 2 more answers

The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was

Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

$[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}$

$[Ca^{2+}]=0.007118 \ M$

$[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}$

$=0.001994 \ M$

Mass of Ca²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}$

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}$

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0

3 years ago

How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2

Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>

4 0

3 years ago