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lina2011 [118]
3 years ago
6

Bombardment of cobalt-59 with a neutron produces a manganese-56 atom and another particle. What is this particle?A. an alpha par

ticleB. a beta particle (electron)C. a positronD. a gamma rayE. a neutron

Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:A

Explanation:

A nuclear reaction is balanced by ensuring that the Masses and charges of te reactants and products are exactly balanced on the left and right hand side of the reaction equation. If there are 60 mass units on the LHS and manganese has only 56 mass units then four mass units are left. If there is no charge on a neutron and there is a charge of 27 on the cobalt, then two charges are left. Four mass units and a charge of +2 corresponds to a helium which is actually an alpha particle.

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Consider the reaction: 2A(g)+B(g)→3C(g). When A is changing at a rate of -0.110M⋅s−1, How fast is C increasing?
mr_godi [17]
From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
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3 years ago
A scientist has 0.12 miles of a gas at a pressure of 4.06 kPa and a volume of 14 liters in an enclosed container. What is the te
natka813 [3]

Answer:

56972.17K

Explanation:

P = 4.06kPa = 4.06×10³Pa

V = 14L

n = 0.12 moles

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T = ?

We need ideal gas equation to solve this question

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume the gas occupies

n = number of moles

R = ideal gas constant

T = temperature of the gas

PV = nRT

T = PV / nR

T = (4.06×10³ × 14) / (0.12 × 8.314)

T = 56840 / 0.99768

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Note : we have a large number for temperature because we converted the value of pressure from kPa to Pa

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8 0
3 years ago
Read 2 more answers
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
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