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3 years ago

Aaleyah learned that astronomers use different properties, such as their

Chemistry

1 answer:

tino4ka555 [31]3 years ago

5 0

Answer:

i need help with that too

Explanation:

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How do you convert between the mass and the number of moles of a substance?

Ksivusya [100]

To convert from grams to moles, you divide the number of grams by the molar mass To convert from moles to grams, you multiply by the molar mass. You must first calculate the molar mass of the substance

6 0

3 years ago

What is the amount in grams of EDTA needed to make 315.1 mL of a 0.05 M EDTA solution. The molar mass of EDTA is 374 g/mol

statuscvo [17]

Answer:

58.92 g EDTA

Explanation:

315.1 mL = .3151 L

M = Moles / Liter

.3151 L x 0.5 mol EDTA x 374 g EDTA = 58.92 g EDTA

1 L EDTA 1 mol EDTA

4 0

2 years ago

A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?

Inessa05 [86]

Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a gas, therefore we can use the ideal gas law:

$PV=nRT$

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 $\frac{atm*L}{mol*K}$

n= ?

So, we can solve for "n" (moles):

$1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K$

$n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}$

$n=0.126~mol$

Now, we have to remember that the molar mass value has "g/mol" units. We already have the grams (9.33 g), so we have to divide by the moles:

$molar~mass=\frac{9.33~grams}{0.126~mol}$

$molar~mass=73.88\frac{grams}{mol}$

7 0

3 years ago

4) How many grams are there in 7.40 moles of AgNO3

Ludmilka [50]

Answer:

1260 grams

Explanation:

3 0

3 years ago

After balancing the following reaction under acidic conditions, how many mole equivalents of water are required and on which sid

nordsb [41]

Answer:

d. 8 moles of H2O on the product side

Explanation:

Hello,

In this case, we need to balance the given redox reaction in acidic media as shown below:

$MnO_4^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq) + Cl^{1-} (aq) \rightarrow Mn^{2+} (aq) + Cl_2 (g)\\\\\\\\(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O\\\\2Cl^{1-}\rightarrow Cl_2^0+2e^-\\\\2*[(Mn^{7+}O^{2-}_4)^{1-} (aq)+8H^++5e^- \rightarrow Mn^{2+}+4H_2O]\\\\5*[2Cl^{1-}\rightarrow Cl_2^0+2e^-]\\\\\\\\2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O\\\\10Cl^{1-}\rightarrow 5Cl_2^0+10e^-\\$

Then, we add the half reactions:

$2(Mn^{7+}O^{2-}_4)^{1-} (aq)+16H^++10Cl^{1-} \rightarrow 2Mn^{2+}+8H_2O+5Cl_2^0$

Thereby, we can see d. 8 moles of H2O on the product side.

Best regards.

4 0

3 years ago