All categories

3 years ago

Can someone please help me with this one

Chemistry

1 answer:

Vinvika [58]3 years ago

4 0

11,835.76 = 11.83576kg

683,763.88mg = 6.8376388hg

53.7m = 0.0537km

:) hoped this helped

brainlyest maybe?

You might be interested in

Please help

Mazyrski [523]

Answer:

Diagram Z

Explanation:

A cell placed into a hypotonic solution will swell and expand until it eventually burst through a process known as cytolysis.

4 0

2 years ago

Compound is the fundamental type of matter that has unique set of properties and cannot be broken down into simpler form.

lubasha [3.4K]

Answer: Compound

Explanation: pure substance that cannot be broken down into simpler substances by chemical or physical means. Each element has unique physical and chemical properties.

6 0

3 years ago

Name one material needed for<br> photosynthesis?

kompoz [17]

One of the materials needed for photosynthesis is sunlight or water

3 0

3 years ago

Read 2 more answers

"The truth of this world" *sigh* What could it be?

dlinn [17]

Answer: "The Truth of This world" in this relation can be any different of anything negative or positive especially with lack of good environment creates tension and stress on the human body both mentally and physically and can cause anger, frustration, hurt, or loneiless.

Explanation:

7 0

2 years ago

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

bagirrra123 [75]

Answer:

$m_{PbI_2}=18.2gKI$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:

$n_{Pb(NO_3)_2}=96.7mL*\frac{1.134g}{mL}*\frac{0.14gPb(NO_3)_2}{1g}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} =0.0464molPb(NO_3)_2\\\\n_{KI}=99.8mL*\frac{1.093g}{mL}*\frac{0.12gKI}{1g}*\frac{1molKI}{166gKI} =0.0789molKI$

Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):

$n_{KI}^{consumed\ by\ Pb(NO_3)_2}=0.0464molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0928molKI$

Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:

$m_{PbI_2}=0.0789molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gKI$

Best regards.

3 0

4 years ago